Linear Programming Duality: An Introduction to Oriented Matroids

A. Bachem W. Kern Linear Programming Duality An Introduction to Oriented Matroids Springer-Verlag Achim Bachem Walte...

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A. Bachem W. Kern

Linear Programming Duality An Introduction to Oriented Matroids

Springer-Verlag

Achim Bachem Walter Kern

Linear Programming Duality An Introduction to Oriented Matroids

With 40 Figures

Springer-Verlag Berlin Heidelberg New York London Paris Tokyo Hong Kong Barcelona Budapest

Achim Bachem Mathematisches Institut, Universitat zu Koln, Weyertal 86-90, W-5000 Koln 41, Fed. Rep. of Germany

Walter Kern Department of Applied Mathematics, University of Twente, P.O. Box 217, NL-7500 AE Enschede, The Netherlands

Mathematics Subject Classification (1991): 05 B 35, 90 C 05, 90C27, 90C48

ISBN 3-540-55417-3 Springer-Verlag Berlin Heidelberg New York ISBN 0-387-55417-3 Springer-Verlag NewYork Berlin Heidelberg Library of Congress Cataloging-in-Publication Data

Bachem, A. Linear programming duality: an introduction to oriented matroids / Achim Bachem, Walter Kern. p. cm. - (Universitext) Includes bibliographical references and index. ISBN 3-540-55417-3 (alk. paper). - ISBN 0-387-55417-3 (U.S.: alk. paper) 1. Matroids. 2. Linear . II. Title. QA166.6.B33 1992 511'.6-dc20 92-14021

programming. I. Kern, Walter, 1957-

This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms orin otherway, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9,1965, in its current version, and a copyright fee must

always be obtained from Springer-Verlag. Violations are liable for prosecution under the German Copyright Law. © Springer-Verlag Berlin Heidelberg 1992 Printed in Germany

Typesetting: Camera ready by author 41/3140-543210 - Printed on acid-free paper

Preface

The main theorem of Linear Programming Duality, relating a "primal" Linear Programming problem to its "dual" and vice versa, can be seen as a statement about sign patterns of vectors in complementary subspaces of JZn.

This observation, first made by R.T. Rockafellar in the late sixties, led to the introduction of certain systems of sign vectors, called "oriented matroids". Indeed, when oriented matroids came into being in the early seventies, one of the main issues was to study the fundamental principles underlying Linear Programming Duality in this abstract setting. In the present book we tried to follow this approach, i.e., rather than starting out from ordinary (unoriented) matroid theory, we preferred to develop oriented matroids directly as appropriate abstractions of linear subspaces. Thus, the way we introduce oriented matroids makes clear that these structures axe the most general - and hence, the most simple - ones in which Linear Programming Duality

results can be stated and proved. We hope that this helps to get a better understanding of LP-Duality for those who have learned about it before and a good introduction for those who have not. The present book arose from an early draft on polyhedral theory and two graduate courses held at Koln university in 1989/1990. No specific prerequisits are assumed. Basically, all we require is some familarity with linear algebra. The book is intended as an introduction to oriented matroid theory for the non-expert. Therefore we restricted ourselves to a thorough discussion of rather elementary results such as linear duality theorems and abstract polyhedral theory.

The only more advanced topic we treat is the topological realization of oriented matroids due to J.Folkman and J.Lawrence, resp. J.Edmonds and A.Mandel. A further reason for restricting ourselves in this way was the fact that at about the same time as this book will appear, a much more comprehensive treatment of oriented matroids by A.Bjorner, M. Las Vergnas, B.Sturmfels, N.White and G.Ziegler including an extensive discussion of more advanced topics will be avail-

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able. Instead of repeating a reference to this work at the end of each chapter of our book we would like to recommend it here, once and for all, to the interested reader. We are grateful to several colleagues for many stimulating discussions on the topic of this book. We are particularly grateful to Michael Hofineister and Winfried Hochstattler who read the "beta version" of the text and provided extensive comments and suggestions. We have worked on this book and written the text at various universities. We acknowledge, in particular, the support of the German Research Association (DFG), the universities of Bonn, Koln and Twente. Our special thanks are due to Frank Neuhaus for the efficient and careful typing the text in TEX.

Koln, October 1991

Achim Bachem Walter Kern

Contents

1

Prerequisites

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1.1 1.2 1.3

Sets and Relations Linear Algebra . . Topology

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2 Linear Duality in Graphs

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2.1 2.2 2.3

Some Definitions . . . . . . . . . . FARKAS' Lemma for Graphs Subspaces Associated with Graphs

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2.4 2.5

Planar Graphs . Further Reading

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3 Linear Duality and Optimization 3.1 3.2 3.3

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Optimization Problems . Recognizing Optimal Solutions Further Reading . . . . . . . .

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4 The FARKAS Lemma 4.1 A first version 4.2 4.3 4.4 4.5 4.6

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Sign Vectors Minors . . . . . . . Oriented Matroids

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57 57 61 63 66 67 70

CONTENTS

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The Composition Theorem . . . Elimination Axioms . . . . . . . Approximation Axioms . . . . . 5.10 Proof of FARKAS' Lemma in OMs 5.11 Duality . . . . . . . . . . . . . . 5.12 Further Reading . . . . . . . . . 5.7 5.8 5.9

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6 Linear Programming Duality 6.1 6.2 6.3 6.4

The Dual Program . . . . . The Combinatorial Problem Network Programming . . . Further Reading . . . . . . MINKOWSKI's Theorem

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Polarity . . . . . . . . . . Faces of Polyhedral Cones Faces and Interior Points The Canonical Map . . . Lattices . . . . . . . . . . 7.7 Face Lattices of Polars . . 7.8 General Polyhedra . . . . 7.9 Further Reading . . . . .

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9 Topological Realizations Linear Sphere Systems

9.2 A Nonlinear OM 9.3

Sphere Systems

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Simplifications . . . Basic Results . . . . Shellability of Topes

8.4 Constructibility of 0 8.5 Further Reading . .

9.1

89 94 107 110

113

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8 The Poset (0, - 1,x2 > 1} is lineal P= {x I xl =X2 =01.

Chapter 2 Linear Duality in Graphs Linear duality deals with the relationship between two complemen-

tary orthogonal subspaces L and L1 of K". The main theorem of linear duality, FARKAS' Lemma, will be presented in Chapter 4. In this chapter we will derive FARKAS' Lemma only for a special class of

complementary pairs (L, Ll) arising from directed graphs.

2.1

Some Definitions

Let V be a finite set, and let E C V x V be a finite family of ordered

pairs e = (u, v). Then G = (V, E) is called a directed graph (or digraph, for short). The elements of V are called the vertices of G, and the elements of E are called the edges of G. Sometimes it is easier just to "draw" a graph rather than describing it explicitely by listing all its edges. For example, if V = {1,. .. , 6} and E = {(1, 2), (1, 5), (1, 5), (1, 6), (2, 4), (2, 4), (2, 6), (3, 2), (3, 5), (4, 5), (5, 5),

(5,5), (6, 5)}, then G = (V, E) may be drawn as indicated below (cf. figure 2.1).

Remark 2.1 The term "directed" is due to the fact that edges are defined to be ordered pairs e = (u, v) E V X V. If one instead defines

an edge to be an "unordered pair" e = {u, v} with u, v E V (with u = v not excluded), one arrives at the definition of graphs. We will always work with digraphs exclusively and therefore sometimes

use the term "graph" as a synonym for "directed graph", since no misunderstanding is possible.

This section is, of course, not considered to be an introduction into graph theory. The interested reader may consult one of the numerous textbooks on graph theory for that. In our context, graphs will only serve as a field where general (abstract) results from linear

CHAPTER 2. LINEAR DUALITY IN GRAPHS

20

1

4

Figure 2.1

duality theory may be interpreted in a more intuitively appealing way. Therefore, we will try to get along with as few definitions as possible, introducing only those concepts which are necessary for developing linear duality theory in graphs.

Let G = (V, E) be a graph. An edge e = (u, v) E E is said to join u to v, and u and v are called the end vertices of e. More precisely, v and u are called the head and tail of e = (u, v), resp. Two vertices are also called adjacent if they are joined by an edge. We also say that u and v are incident with e = (u, v) and vice versa. Two edges e and e' are called incident if they have at least one end vertex in common. In case e and e' have two end vertices in common, they are called parallel or antiparallel, according to whether the tail of e equals the tail of e' or the head of e'. An edge with just one end vertex is called a loop. A graph without loops and (anti-) parallel edges is called simple. We say that G= (V', E') is a subgraph of G = (V, E) if V' C V and E' C E. If G' contains all edges of G whose end vertices are in

v,, then G' is said to be the subgraph induced by V', denoted by G[V'].

2.1. SOME DEFINITIONS

21

We shall often construct new graphs from old ones by "deleting" or "contracting" edges. If e E E, then G:= G\e := (V, E\e) is said to be obtained from G by deleting the edge e. More generally, if A C E, then G \A denotes the graph which is obtained from G by deleting all edges e E A (successively). If e = (u, v) E E, then G' := Gl a denotes the graph which is obtained from G by identifying the vertices u and v and removing the edge e. We say that G/e is obtained from G by contracting the edge e. More generally, if A C E, then G/A denotes the graph obtained from G contracting all edges e E A (successively). It is easily seen that these two operations commute, i.e. if A and B are two disjoint subsets of E, then G \ A/B = G/B \ A. The graph G \ A/B is called a minor of G.

e

EI

f

G

f

e

G/f

G\e

G\e/f

Figure 2.2: Contraction and Deletion A path in G is an alternating sequence p of vertices and edges, say p = (vo, e1, v1, ... , el., vk) such that each e; has endvertices vi-1 and vi,

and the edges are pairwise distinct. (This latter restriction is usually not made in the definition of paths.) If p = (vo, e1, v1i ... , ek, vk) is a path, then vo and vi, are called its initial and terminal vertex, and

k is called its length. We also say that p is a path from vo to vk. The path p is called simple if the vertices vo, ... , vk_1 are pairwise distinct. The path p is called closed if its initial and terminal vertex coincide. The closed path p is a circuit if the subpath with initial

22


vertex vo and terminal vertex vk_1 is simple. If p = (vo, e1, v1, .... ek, vk) is a path, then e; is called a forward or backward edge, according to whether e; = (vs_I, vi) or (vi, vs_1). Usually, a path is not uniquely determined by its forward and backward edges but this minor ambiguity will not be essential in our context.

Therefore, we often identify a path p with its set P C_ E of edges, partitioned into P+ and P-, the sets of forward and backward edges, resp. Note that if p is a circuit, then P+ and P- are essentially (i.e., up to interchanging them) determined by the whole set P C E of edges of p. In fact, there are only two possible ways (opposite directions) to go along a circuit. A graph G = (V, E) is called connected if any two of its vertices are joined by a path. Otherwise, G is called disconnected. A maximal connected subgraph of G is called a component of G. A subset Y C E of edges is called disconnecting if G \ Y has more components than G. A minimal disconnecting set Y C E is called a cut. For reasons to become clear later in Section 2.3, cuts are also called cocircuits.

If Y C E is a cut, then obviously there exist two induced subgraphs, say G1 = G[VI] and G2 = G[V2] such that Y is the set of edges joining G1 and G2 in G. Conversely, any such set, i.e. any non empty set Y C E, consisting of precisely those edges that link two

induced subgraphs G1 and G2 of G, is a cut. We call such a cut a GI-G2-cut. If u and v are vertices of G1 and G2, resp., then Y will also be called a u-v-cut or a u--rr-cocircuit. If G = (V, E) is a graph and V1 C V, then 8(V )C E denotes the set of edges linking VI to V2 := V \ V1. There is a natural bipartition of 8(VI) into the set 8+(VI), consisting of those edges whose tail is in V1 and whose head is in V2 = V \V1i and the set 8-(V1) consisting of those edges whose tail is in V2 = V \ V1 and whose head is in V1. By our above consideration, a cocircuit in G may be respresented by a set Y = 8(V1) C E, partitioned into Y+ = 8+(VI) and Y- = 8-(V1). Again, Y+ and Y- are called the set of forward and backward edges of Y, resp. Note that, as in the case of circuits, if Y is a cocircuit, then Y+ and Y- are essentially (i.e., up to interchanging them) uniquely determined.

2.2

FARKAS' Lemma for Graphs

If you consider a given graph G = (V, E) as a system of "one way streets", then a u-v-path P with P- = 0 represents a feasible way to go from u to v. Now suppose someone (say, your boss) hands you a graph, together with two prescribed vertices u and v, and asks you

2.2. FARKAS' LEMMA FOR GRAPHS

23

to find out whether such a path P from u to v with P- = 0 exists. He does not consider you to be a very trustworthy person, so he will

not be content with getting just a "yes" or "no" answer, but he in addition wants you to prove that your answer is correct. Obviously, in case such a path exists, you will eventually find it by searching the graph in any systematic way. (The most simple and

stupid way would be to list all subsets P C E and check them one by one.) Once you have found a path as required, you may hand it to your boss as a proof, showing that your "yes" answer is correct. In case you find that no such path exists, is there a simple way to convince your boss that your "no" answer is correct? In fact, there is: Suppose that during your examination of the graph, you have found

a u-v-cut Y C E, dissecting G into components G1 and G2 such that G1 contains u and G2 contains v and all edges in between are directed from G2 to G1, i.e. Y is a u-v-cut with Y+ = 0. This gives evidence to the fact that no u-v-path P with P- = 0 can exist. The following theorem states that such a cut always exists, provided your

"no" answer is correct, i.e. in case there is no path from u to v as required.

Theorem 2.2 (FARKAS' Lemma for Graphs) Let G = (V, E) be a connected graph, and let u, v E V. Then either

a) 3 u-v-path P with P- = 0. or

b) 3 u-v-cut Y with Y+ = 0. but not both

.

Proof. The proof is constructive, i.e. we will explicitely describe an

algorithm which either finds a path as in a) or a cut as in b). The algorithm proceeds by constructing the set V1 C V of all vertices that can be reached by a directed path starting at u. Initially, the set V1 contains only u itself.

Vi:={u} WHILE 8+(V) 96 0 and v 0 V1 DO BEGIN

let e = (s, t) E 8+(V1), set Vi := Vi U {t}, and label(t) := s END.


24

Obviously, the algorithm stops after at most IVI steps. If it ends up with v E V1, we know that there exist a directed path from u to v. This may be traced back from v by means of the labels. If the algorithm stops with v 0 V1, then 5(V1) = 8-(V) is a cut as in b). This simple result is the main theorem of linear duality in the con-

text of graphs. Note that the general idea is to replace the statement "there exists no ..." by an equivalent statement "there exists...". This general aspect of linear duality will be emphasized also in Chapter 3 where we investigate linear duality in the context of optimization. Let us conclude this section by restating the above theorem in a slightly different version.

Corollary 2.3 (FARKAS's Lemma) Let G = (V, E) be a graph. Then for every e E E either

a) 3 circuit X C E such that e E X and X-= 0. or b) 3 cocircuit Y C E such that e E Y and Y+ = 0. but not both .

Proof. Let e = (v, u) and apply Theorem 2.2 to the graph G \ e. Obviously, a circuit containing e is made up of e and a path from u to V. 0

2.3

Subspaces Associated with Graphs

In this section we will associate two complementary subspaces L and Ll to a given graph and see how the duality results from Section 2.2

may be restated in terms of L and Ll.

Definition 2.4 Let G = (V, E) be a graph. Then 1KE is called the edge space of G. (Recall from Section 1.2 that we consider the elements of 1KE as vectors with coordinates indexed by E, rather than

functions from E to IK.) If X C E is a circuit, partitioned into X+ and X- (the sets of forward and backward edges), then x E IKE, defined by

1+1 ifeEX+ Xe

-1 ifeEX0 ifeEX

is called the incidence vector of X.

(e E E)

2.3. SUBSPACES ASSOCIATED WITH GRAPHS

25

Similarily, if Y C E is a cocircuit, partitioned into Y+ and Y-, then y E 1KE, defined by

1+1 ifeEY+

-1 ifeEY0 ifeEY

Y.=

(eEE)

is called the incidence vector of Y. (In the following, circuits and cocircuits will usually be denoted by upper case letters, U, X, Y, Z... and their corresponding incidence vectors in IKE will be denoted by lower case letters u, x, y, z....) The circuit space of G is the subspace of 1KE which is generated by the incidence vectors of the circuits of G. Similarily, the cocircuit space of G is the subspace of ]KE which is generated by the incidence vectors of the cocircuits of G.

Example 2.5 Consider the graph drawn below. e6

e1

f

e2

T

e5

e7

a e3

e4

The edge space of the above graph is 1KE _- N. The circuit made up of X+ = lei, e2, e7} and X- = {e3} corresponds to the incidence vector

x=(1,1,-1)0,0,0,1). Similarily, we see that

x'=(0,0,0,-1,1,1,1) is the incidence vector of a circuit. Hence, for example,

x + x'= (1, 1, -1, -1,1,1, 2) is an element of the circuit space. The two vectors y = (0, 0, 1, 0, -1, 0,1)

y' = (-1'0'-l'0'0'0'0)

and

CHAPTER 2. LINEAR D UALITY IN GRAPHS

26

are incidence vectors of cocircuits. Hence, for example,

y +3y'= (-1,0,-2,0,-1,0,1) is an element of the cocircuit space. In the following we are going to show that the circuit and cocircuit space of a graph form a complementary pair L, Ll of subspaces in KE.

Lemma 2.6 The circuit- and cocimuit space of a graph G = (V, E) are orthogonal.

Proof. Let X C E be a circuit, partitioned into X+ (its set of forward edges) and X- (its set of backward edges). Let Y C E be a cocircuit, dissecting G into Gl and G2 and assume that Y+ is the set of edges directed from Gl to G2, and Y- is the set of edges directed from G2 to G1. Let x and y, resp. denote the corresponding sign vectors. Now it is easily seen that the inner product xT y equals the number of times we go from Gl to G2 minus the number of times we go from G2 to G1, when following the circuit X. This is obviously equal to zero. Thus x and y are orthogonal. Since the circuit and cocircuit space of a graph are generated by circuits and cocircuits, resp., this proves the claim. 0

To prove that the circuit- and cocircuit space of a graph are in fact complementary, we have to investigate their dimensions. For that purpose, we introduce the following notions:

Definition 2.7 Let G = (V, E) be a graph. A set A C E is called independent if it contains no circuit X C A. A maximal (with respect to set inclusion) independent subset A C E is called a basis of G. Example 2.8 A basis of a graph with two components


27

Lemma 2.9 Let G = (V, E) be a graph with n vertices and k components. Then every basis of G has precisely n - k elements.

Proof. The proof is by induction on n. If n = k, i.e. G consists of n isolated vertices, the claim is trivially true. Thus assume n > k, and let A be a basis of G. Choose any edge e E A, and consider G' := G/e. Obviously, G' again consists of precisely k components. Furthermore, it is immediate that A \ e is a basis of G'. In fact, every circuit in G' either corresponds to a circuit in G or to a path between the two end

vertices of e. Hence, by induction, IA \ e = n - 1 - k, which proves the claim. 0

Lemma 2.10 Let G = (V, E) be a graph with n vertices, m edges and k components. Then the dimension of its circuit space is at least m - n + k, and the dimension of its cocircuit space is at least n - k.

Proof. Assume first that G is connected, i.e. k = 1. Let A C E be a basis of G. Then any two vertices of G are linked by a unique path P C A. (In fact, such a path exists since (V, A) is connected, and it is unique since A does not contain a circuit.) For every e E E \ A let X(e) denote the circuit of G that is made up of the unique path P C_ A linking the two end vertices of e, and the edge e itself, say as a forward edge. Let x(e) E KE denote the corresponding incidence vectors. These form a set of I E \ Al = m - n + 1 vectors in the circuit space of G which is linearly independent. In fact, for each particular e E E \ A, the vector x(e) has a "+1" entry in coordinate e while all other vectors have coordinate e equal to 0. Thus the dimension of the circuit space is at least m - n + 1. Next consider the cocircuits. If e E A, then G' = (V, A\ e) consists of two components, say Gl and G2. Let Y(e) denote the cocircuit of G consisting of those edges joining Gl and G2 in G. These cocircuits give rise to a set of IAA = n-1 independent incidence vectors y(e), e E A, in the cocircuit space of G. In fact, for a given e E A, y(e) has a nonzero coordinate e, while all the others have a zero entry in coordinate e. Thus the cocircuit space of G has dimension at least n - 1. The proof is finished for k = 1. In case G has more components, the result follows easily by choosing independent circuits and cocircuits in each component. 0 The last two lemmata may be combined to obtain the following fundamental result:

CHAPTER 2. LINEAR DUALITY IV GRAPHS

28

Theorem 2.11 The circuit- and cocircuit space of a graph G = (V, E) form a complementary pair L, L1 < IKE. 0

The linear duality theorem we proved in Section 2.2 may thus be restated as follows:

Theorem 2.12 (FARKAS' Lemma) Let L, L1 _< IKE denote the circuit- and cocircuit space of a graph G = (V, E). Then for every e E E either

a) 3 x E L ,eEsuppx,x>0 or

b) 3 y E L1, e E supp y, y< 0 but not both. 0

(Note that the "either - or" part follows from the corresponding version of these two theorems in Section 2.2. The appendix "but not both" is trivial since if x is as in a) and y as in b), then xTy < 0, i.e., x and y are not orthogonal.) The general principle behind is the main theorem of linear duality, stating that the above result remains valid if L is an arbitrary subspace of KE. This will be proved in Chapter 4. Maybe, the difference between subspaces associated to graphs in the above way and general subspaces of KE is not very clear at this point. In the following we will represent the circuit- and cocircuit spaces of graphs in a way which makes their special structure more evident. We know from linear algebra (cf. Section 1.2) that every complementary pair of subspaces L, L1 in KE can be represented by a

matrix A such that L = ker A and L1 = im AT. In the case of "graphical" subspaces there is a canonical choice for the matrix A:

Let G = (V, E) with V = {vl,... , v"} and E = {el,... , em} without loops. Let A = (ai,,) E K` be defined as +1 if vi is the head of e..,

ai; _

-1 if vi is the tail of e 0 if vi and e1 are not incident.


29

This matrix A is called the incidence matrix of G. Thus each edge ej = (vi, vk) of G is represented by a column

10\ 0

-1

+- i

0

A.; _ 0

4- k

1

0

0) One may also define incidence matrices for graphs containing loops by adding a zero column for each loop. Each vertex v; of G is represented by a row A;. which has a "1" entry for every edge that has v; as its head and a "-1" entry for every edge that has v; as its tail.

Example 2.13 Consider the graph G = (V, E) as drawn below.

Its incidence matrix is given by

-1

A=

0

1

-1

0 0 0

0 0

1

0 -1 0 0 0 0 -1 0 -1 0 0 -1

0 0 0

0

0

1

1

-1

1

1

0

0

1

0 0 0 0 0

0 0 0 0 0

30

CHAPTER 2. LEVEAR DUALITY EV GRAPHS

Lemma 2.14 Let G = (V, E) be a graph, and let A denote its incidence matrix. Then the circuit space of G is contained in ker A. Proof. Let P = (vo, e1) v11 ... , e,, vk) be a simple path. Consider its

incidence vector p E KE which has a "+1" entry for each forward edge of P and a "-1" entry for each backward edge of P. Then it is obvious by induction on k that if vo # vk, then /

ON 0

-1

4- vo

0

Ap = 0 1

4- vk

0

0) From this it is immediate that if X C E is a circuit (made up of a path from vo to vk and an edge joining vo and vk) and x E KE is its incidence vector, then Ax = 0. This proves the claim. D

Lemma 2.15 Let G = (V, E) be a graph, and let A denote its incidence matrix. Then the cocircuit space of G is contained in im AT. Proof. Let Y C E be a cocircuit, and let V = V1LJV2 be a corresponding partition of V. Say that Y+ is the set of edges directed from Vl to V2, and Y- is the set of edges directed from V2 to V1. Let u E KV be the incidence vector of V1, i.e. u has a "1" entry for every v E Vl and a "0" entry for every v E V2. Then it is easily seen that the incidence

vector y EKE of the cocircuit Y can be written as

y=uTA, i.e., y E im AT. D

Theorem 2.16 Let G = (V, E) be a graph with incidence matrix A. Then the circuit- and cocircuit space of G are ker A and im A, T resp..


31

Proof. This is immediate from the two lemmata above and the fact that the circuit- and cocircuit space of a graph are complementary. 0 This result tells us how to get the circuit- and cocircuit space of a given graph G = (V, E). What about the converse problem, i.e. suppose we are given a subspace L < KE, say in terms of a matrix A such that L = ker A, and we want to reconstruct the associated graph q (provided there is one)? The easiest way to do this, is to transform A into an incidence matrix by means of elementary row operations (which do not affect the null space ker A). If one does not succeed, then L is not the circuit space of a graph. If one succeeds, the desired graph G is given by the incidence matrix A. The following result tells us how the circuits and cocircuits of G may be obtained from the given circuit- and cocircuit spaces L, L1 directly:

Proposition 2.17 Let G = (V, E) be a graph, and let L, L1 < IKE denote the circuit- and cocircuit spaces of G, resp. Then a vector x E IKE is an elementary vector of L if and only if it is a scalar multiple

of a circuit incidence vector of G. Similarily, a vector y E IKE is an elementary vector of L1 if and only if it is a scalar multiple of a cocircuit incidence vector of G.

Proof. Let A denote the incidence matrix of G. We claim that J C E is independent if and only if A. j consists of linearly independent columns. In fact, A.j is the incidence incidence matrix of G' = (V, J). Hence J C E is independent if and only if it contains no circuits, i.e. if and only if the circuit space of G', which is ker A.j, is zero. This is equivalent to saying that the columns of A.j are linearly independent.

Now let x E KE. If X E elemL, then every proper subset J C suppx is independent in G. But X = suppx itself is dependent. This shows that X itself is a circuit of G. Since elementary vectors having the same support can differ only by a scalar multiple, the claim follows.

Next let us turn to the cocircuits. Let z = uTA be an elementary vector of the cocircuit space of G. Suppose that G consists of, say, k components, induced by V1, .

Vk C_ V. If u, considered as a map V - K, were constant on each component then obviously z = 0, i.e. z is not an elementary vector. Hence, for some V, there is a nontrivial partion V, = U Vi" such that u is constant, say u = 1, on .

.,

and u # 1 on each vertex of V". But this is obviously equivalent to saying that suppz contains all edges joining to Vi". Thus suppz contains a cocircuit Y C E and since z is elementary, this implies that supp z = Y. Hence every elementary vector in im AT is a scalar multiple of a cocircuit incidence vector. That cocircuit incidence vectors

32


in turn are elementary vectors in im AT can be seen by essentially reversing the above argument. In fact, let Y C E be a cocircuit, and let y E KE denote its incidence vector. If V, = U and V", then y can be V, such that Y consist of all edges joining written as y = uTA with u - 1 on Vi', u - 0 on Vi" and u taking some constant value on each other V. If z = wTA is any nonzero vector with supp z C supp y then w must also take constant values on each V j # i. Furthermore, w must take a single constant value, say A' on But this implies already and a single constant value A" $ A' on supp z D supp y. Hence y is in fact elementary. D

2.4

Planar Graphs

This section may be considered as an appendix to Chapter 2 and therefore we will ommit some of the proofs. Our main object is to introduce a special class of graphs ("planar graphs"), where duality between circuits and cocircuits becomes most evident (or "visible").

A graph G = (V, E) is called a planar graph if it can be drawn in the plane in such a way that no two edges cross each other. To put it a little bit more precisely, one may say that the edges are allowed to be simple JORDAN curves connecting the points corresponding to the end vertices of an edge and that by "non crossing" we mean that two edges connecting different vertices have no point in common except possibly one of their endpoints. One can show that this is equivalent to requiring every edge to be represented by a straight line segment, one for each class of (anti-) parallel edges and disregarding loops. Such a representation of G will be called a plane graph. There is a wellknown theorem, due to KURATOWSKI which characterizes planar

graphs. This states that a graph G = (V, E) is planar if and only if G has no minor "isomorphic" to one of the following two graphs

2.4. PLANAR GRAPHS

33

Known proofs of this result are rather long (though elementary) and therefore ommitted here. Now consider a plane graph G = (V, E). Removing its vertices and edges from the plane leaves a number of connected components, called faces or countries. Clearly each plane graph has exactly one

unbounded face. The boundary of a face is the set of edges in its closure. Two countries are called neighbouring if their boundaries have an edge in common. Let us mention the following result, known as EULER's formula, which we shall meet again in a more general context in Chapter 9.

Theorem 2.18 (EULER'S formula) Let G = (V, E) be a connected plane graph with n vertices, m edges and f faces. Then

n-m+f =2. Proof. We apply induction on the number of faces. If f = 1, then G has no circuits, i.e. E is a basis of G and hence m = n - 1 so the result holds. Now suppose that f > 1. Then there is a bounded face whose boundary obviously is a circuit X C E of G. Hence deleting an edge e E E leaves the graph connected and decreases both f and m by 1. Thus the formula follows by induction.

The most important thing about planar graphs in our context is the concept of dual graphs. This is as follows. Suppose G = (V, E) is a plane graph. Then we may define a new graph by considering each country as a vertex and by joining two vertices by an edge, provided the corresponding countries are neighbouring. More precisely, assume c is a bounded country, and let P C E be the circuit consisting of the

boundary of c such that P+ and P- are the forward and backward

CHAPTER 2. LINEAR DUALITY IV GRAPHS

34

edges if we go around P in a counterclockwise orientation. Then for every edge e E P we include an edge e' in the dual graph joining the country c to the neighbouring country c' which is separated from c by e. More precisely, if e E P+, the edge e' will be directed from c to c'

and if e E P-, the edge e` will be directed from c' to c. The graph obtained this way is called the dual of G. This is obviously a planar graph again.

Example 2.19 A planar graph and its dual (represented by dotted lines) ----------------

If G = (V, E) is a planar graph, its dual will be denoted by Gl = E'). Thus if n, m and f denote the number of vertices, edges and faces of G and n', m* and f* are the corresponding numbers of Gl, a then m = m", n = f " and f = n symmetric role in EULER's formula.

If X C E is a circuit of G, then X is the boundary of the union of faces it encludes. The corresponding set X' C E` consists of those edges in E* that join a country "within" X to a country "outside" X. This shows that every circuit of G corresponds to a cocircuit of Gl. (The details are left to the reader.) If G is connected, its circuit space

has dimension m - n +1 = f -1 = n' -1, which is the dimension of the cocircuit space of G1. Hence, by the observation above, the circuit space of G is equal to the cocircuit space Gl. (More precisely, the two spaces are isomorphic, the isomorphism being induced by the obvious

map KE - KE'.) This further implies that the cocircuit space of G is the circuitspace of G1. Since circuits and cocircuits can be identified as elementary vectors within the spaces they generate, this gives a one-to-one-correspondence between circuits of G and cocircuits of G.

2.5. FURTHER READING

35

Theorem 2.20 Let G = (V, E) be a connected planar graph, and let G* = (V*, E*). Then X C E is a circuit of G if and only if X* C_ E* is a cocircuit of G*, and Y C E is a cocircuit of G if and only if Y* C E* is a circuit of G*.

2.5

Further Reading

Classical textbooks on graph theory are: C. BERGE The Theory of Graphs and its Application, J. Wiley, New York (1962).

F. HARARY Graph Theory, Addison-Wesley, Reading, Mass. (1969).

Those who are interested in the history of graphs, may consult the book "Graph Theory 1736-1936" by N.L. BIGGS, E.K. LLOYD and R.J. WILSON, Oxford University Press, London (1976). Since 1960, the field has seen a real explosion of research and applications. We therefore list just a very few out of the huge number of more recent textbooks on graph theory: B. BOLLOBAS Graph Theory, Springer Heidelberg (1979).

C. BERGE Graphs, North-Holland, Amsterdam (1985).

M. GOUDRAN, M. MINOUX and S. VAJDA Graphs and Algorithms, J. Wiley, New York (1984).

W. T. TUTTE Graph Theory, Encyclopedia of Mathematics and its applications, Vol 21, Addison Wesley, Reading, Mass. (1984).

Chapter 3 Linear Duality and Optimization 3.1

Optimization Problems

Many problems in the real world (and in mathematics) can be formulated in the following way:

Let f : K" -+ K be a function, and let S C W. Find x E S such that f (x) > f (x) for all x E S.

Such a problem is called a mathematical programming problem (mathematical optimization problem) or simply a mathematical program. More precisely, one may call it a maximization problem, because we are looking for ant E S that maximizes the function f . Analogously, in case we seek for an x E S with f (x) < f (x) for

all x E S we speak of a minimization problem. The set S C_ K" is called the constraint set or the set of feasible solutions. The function f : K' --+ K is called the objective function. Any x E S that maximizes (resp. minimizes) f on S is called an optimal solution. It is common to shorten the notation of the statement of an optimization problem by writing

maxf(x)

xES although this is a slight misuse of the max operator in general. (In fact, it would be more appropriate to replace it by "sup", since optimal solutions may not exist in general, as discussed below.)

From the above statement of an optimization problem it is not quite clear what it means to "solve" an optimization problem. Let

38

CHAPTER 3. LINEAR DUALITY AND OPTIMIZATION

us examine this question in more detail. Note that optimal solutions may fail to exist for several reasons. First, it may happen that S = 0, i.e. there are no feasible solutions at all. In this case the problem is

called inconsistent. Otherwise, i.e. if S $ 0, the problem is called consistent. Whether or not a given problem is consistent may not be obvious in advance. (As discussed below, it may be arbitrarily hard to find out whether or not S = 0, depending on how S is defined.) Secondly, it may happen that the objective function is unbounded on S, i.e. there exists a sequence x1, x2, ... of feasible solutions x; E S such that f (x;) > i for every i E IN. Obviously, in this case again no optimal solution exists. Finally, even in case the problem is bounded, i.e. there exists an upper bound u E R such that f (x) < u for every

x E S, it may happen that for every x E S there exists a y E S such that f (x) < f (y). (For example, choose S = [0, 7r] fl Q, and let f (x) := x.) In this case the problem is called degenerate. Summarizing, a necessary condition for the existence of optimal solutions is that the problem is consistent, bounded and nondegenerate. Solving an optimization problem now amounts to find out and prove whether the problem is consistent/bounded/degenerate and in case optimal solutions exist, to find one and prove that it is indeed optimal. Solving an optimization problem may be arbitrarily hard, depending on how complicated the constraint set S or the objective function f are allowed to be. In particular, there are classes of optimization problems that can provably not be solved by any algorithm. (This statement, of course does not mean anything, as long as one does not define precisely what should be understood by the term "algorithm". We do not want to give a rigorous definition here because this subject is outside the scope of our book. The interested reader is asked to consult any book on computability for more about this topic.) The main reason for the unsolvability of general optimization problems

is GoDEL's incompleteness theorem and the work arising from HILBERT's tenth problem, cf. the references given in Section 3.3. Consider for example the following class of problems. For each polynomial p(x) = p(x1i ... , x") with integer coefficients in n variables x1,. .. , x" let f (x) :_ lp(x)l be the corresponding objective function, and let S:= 7L" be the constraint set. Then solving the optimization problem min f (x)

xES is at least as difficult as deciding wether or not there exists an x E 7L" such that p(x) = 0. One can prove that there is no algorithm

3.1. OPTIMIZATION PROBLEMS

39

which, given an arbitrary polynomial p as input, decides whether or not p(x) = 0 has a solution x E 7L". Thus one has to impose some restrictions on the set S C_ K" and the function f : K" - K in order to get classes of solvable problems. One possible restriction, for example, is to choose the constraint set as a convex set S C K" and to allow only concave functions f : K" - K (in case of a maximization problem). The class of problems arising

this way is the class of socalled convex optimization problems. The interested reader is invited to consult any book on mathematical programming in order to learn about how to solve such kinds of problems or other restricted classes of optimization problems. Here we will be concerned only with a still more restrictive class of problems,

the socalled linear optimization problems, which arise by taking f : K" --' K as a linear function, i.e.,

f(x)=cTx for some c E K", and by choosing the constraint set S C K" to be a polyhedron

S=P(A,b)={xEK" I Ax 0.

Proof. This is a straightforward exercise. Let I denote the set of active constraints for x, and let J := {1, ... , m} \ I. Then Aj.x = bI and Aj.x < bj. Thus if z E K", we may choose A > 0 small enough to ensure that Aj.(x + Az) < bJ. Thus y = x + Az E P for sufficiently small A > 0 if and only if Al.y < bi, i.e. Al.z < 0, which means that z is a feasible direction for x. 0

Corollary 3.4 Let P = P(A, b) C IK", let x E P, and let I denote the set of active constraints for x. Let Q := {y I Al.z < 0, cTz = 1}. Then the two alternatives a) and b) from above are respectively equivalent to

a') Q is nonempty. b') Q is empty. Proof. This is immediate from Lemma 3.3. Q is nonempty if and only if there is a feasible direction z satisfying cTz = 1. This in turn


43

is equivalent to saying that for A > 0 sufficiently small there exists a better solution y = x + Az E P.

The linear duality theorem, which will be stated and proved in Chapter 4, says that to every polyhedron Q C If one can efficiently compute a polyhedron Q' C K' such that Q is empty,* Q' is non empty. This shows that (at least in principle) it is possible to decide which of the above alternatives a') and b') holds: We simply choose any enumeration x1, x2i x3... of the rational vectors x E Q" and successively

check whether x; E Q or xi E Q'. In any case, after a finite number of steps, we will eventually discover that Q is nonempty or Q' is nonempty.

3.3

Further Reading

The two most important historical papers on decidability are:

K. GODEL On Formally Undecidable Propositions of Principia Mathematica and Related Systems I, Monatshefte fur Mathematik and Physik 38 (1931), pp. 173-198, reprinted in S. Ferman et al, "Kurt Godel: Collected Works I: Publications 1929 - 1936", Oxford University Press (1986).

A.M. TURING On Computable Numbers with an Application to the Entscheidungsproblem, Proceedings London Mathematical Society 42 (1937), pp. 230-265, reprinted in M. Davis, "The UndecidableBasic Papers on Undecidable Propositions, Unsolvable Problems and Computable Functions", Hewlett: Raven Press (1965).

Known proofs of the undecidability of HILBERT's Tenth Problem are rather involved. The interested reader may consult J.P. JONES and MATIJASEVIC Register Machine Proof of the Theorem on Exponential Diophantine Representation of Enumerable Sets, Journal on Symbolic Logic 49 (1984), pp. 818-829,

presenting a simple proof of a related result and more references on that subject. The following list contains some references on (general) optimization and linear optimization:

44

CHAPTER 3. LINEAR DUALITY AND OPTIMIZATION

V. CHVATAL Linear Programming, W.M. Freeman, New York (1983).

G. B. DANTZIG Linear Programming and Extensions, Princeton University Press, Princeton, N. J. (1963). R. FLETCHER Optimization, Academic Press (1969).

D. LUENBERGER Introduction to Linear and Nonlinear Programming, Addison Wesley (1973).

G. MCCORMICK Nonlinear Programming, John Wiley (1983).

K. MURTY Linear and Combinatorial Programming, John Wiley (1976).

A. SCHRIJVER Theory of Linear and Integer Programmning , John Wiley (1986).

A brief sketch of the history of linear programming is given by the "founder" of linear programming, George Dantzig, in his article "Reminiscenses about the Origin of Linear Programming", in: A. BACHEM, M. GROTSCHEL and B. KORTE Mathematical Programming. The State of the Art, Springer (1983).

Chapter 4 The FARKAS Lemma The result mentioned at the end of Section 3.2, which relates the emptiness of a given polyhedron to the nonemptiness of a certain other polyhedron is wellknown as the FARKAS Lemma. We will state this theorem in a more precise form in this section. Moreover, we shall give several equivalent formulations of the FARKAS Lemma, which we derive from each other by introducing standard techniques in polyhedral theory. In particular, we will show that the FARKAS Lemma can

be seen as a theorem relating a subspace L of K" to its orthogonal complement Ll in the sense of Theorem 2.10.

4.1

A first version

Recall that our problem in Section 3.2 was to verify the emptiness of a polyhedron given by a set of linear inequalities. Suppose P = P(A, b) is an empty polyhedron. Hence assuming Ax < b

for some x E K"

(4.1)

must lead to a contradiction. By monotonicity of addition and multiplication with nonnegative scalars, it is immediate that (4.1) implies

du E K+

uTA2 < uTb.

(4.2)

Since we do not know anything about x, we will in general not be able to decide whether uTAx < uTb holds for a given u E K+ or not. Therefore we are interested in those inequalities which are independent of x, i.e. we consider those u E K+, which satisfy uTA = 0. This leads to

VuEK+

(uTA=0=:,. uTb>0).

(4.3)

CHAPTER 4. THE FARKAS LEMMA

46

Thus if we find u E K+ such that u TA = 0 but uT b < 0, we have a contradiction to (4.1). FARKAS' Lemma now also proves the converse,

i.e. if we can not find u E K+ such that uTA = 0 and UTb < 0, then statement (4.1) must be true.

Theorem 4.1 (FARKAS' Lemma) Given A E 1K"` and b E IK'n, then either

a) 3xE1K",Ax0

but not both.

Proof. We shall only show that Theorem 4.5 and Theorem 4.1 imply

each other. The proof of both statements will be given in Section 4.3. Clearly Theorem 4.5 a) is equivalent to "3 z E Kn+1, Bz < 0, -1" or "3 z E K"+1, Bz < 0, zn+l = 1." Again this can be reformulated as "3 (1) E Kn+1, B(1) _< 0" or as "3 x E Kn : Ax < b" which is 4.1 a). In the same way, we see that 4.5 b) is equivalent to 4.1 b). Theorem 4.5 is just a restatement of Theorem 4.1, and there seems to be nothing interesting about it. However, we are now prepared to change our point of view. Note that the set {Bz I z E Kn+1} = im B

is a vectorspace in K'"+1 and that {u I uTB = 0} = kerBT is its orthogonal complement. Thus the FARKAS Lemma can be seen as a theorem relating pairs of orthogonal vectorspaces. We will emphasize this point of view in the next section.

4.3

Linearization

As we used homogenization to pass from polyhedra to polyhedral cones, we will use "linearization", to pass from polyhedral cones to linear vectorspaces. However, "linearization" is not really a "technique"

in the usual sense, but rather a "change of viewpoint". Suppose we are given a polyhedral cone C = P(B, 0), B E K('"+1)x(n+1) Instead of looking at C C 1K , we may as well apply the transformation B : Kn+1 . K'+1. Anything happening to C in Kn+1 will then correspond to something happening in im B C ICm+1. Any "polyhedral statement" concerning C will correspond to a "linear subspace statement" concerning L = im B. As an example, how we can use "linearization" to translate statements about polyhedral cones into equivalent statements about vectorspaces, we shall restate the FARKAS Lemma as follows:

CHAPTER 4. THE FARKAS LEMMA

50

Theorem 4.6 Let L < 1K"`+1 be a subspace of

1K'+1.

Then either

0,ym+1>0

a) or

b) 3uELl,u>O,um+l>0 but not both.

Proof. We shall first show that Theorem 4.6 is equivalent to Theorem 4.5 and will then give a proof of Theorem 4.6. Given A E Kmxn , b E

K'", and B =

OA

_b

/

as in Theorem 4.5, we set L := im B, thus

L1 = her BT. Now Theorem 4.6 states, that either

a) 3y=Bz,y>0,

ym+l>0

or

b) 3u>0, UTB-O,um+1>0 but not both. This is exactly the content of Theorem 4.5. To show that Theorem 4.6 follows from Theorem 4.5, let L be an arbitrary subspace of 1('"+1 Since L and Ll together span Km+1,

it cannot happen that both L and Ll are contained in K' x {0}. Due to the symmetry of the conditions a) and b) in Theorem 4.6, we may assume w.l.o.g. that L is not contained in K'" x {0}. Hence L can be written as L = im B, where B E K('"+1)x(n+1) has the form A 0

b

-1). The claim now follows immediately from Theorem 4.5.

Proof. (Theorem 4.6) Obviously, the two alternatives a) and b) cannot hold simultaneously. For, if y is as in a) and u is as in b), then yT u > 0, contradicting the definition of Ll. Therefore, it is sufficient to show that for every subspace L < 1(m+1, there exists y E L and u E Ll such that y > 0, u > 0 and ym+1 + um+1 > 0. This is clearly true if m = 0, thus we may proceed by induction on m. Assume that

m > 1, and let A be a matrix such that L = her A. Now delete the first column of A and denote the resulting matrix by A. Then by our induction hypothesis, there exist y = (y2, ... , ym+l) E

kerAand w=(fL2i...,gym+l)=tTAEimAT,such that y>0,ft > 0 and 9m+1 +iim+l > 0. Considering

Ll, we see that

(0, y) E L and u

wTA E

4.4. DELINEARIZATION

(1) There exists y E L, u E L1 such that and y.+1 + um+l > 0.

51

0, (u2i ... , um+1) > 0

By symmetry, our induction hypothesis implies:

(2) There exist y E L, u E L1 such that (y2 ... , y,n+l) > 0 , u > 0, and y,n+1 + um+l > 0

-

If y and a are as in (1) and u1 > 0, we are done. Similarily, if y and u are as in (2) and yl > 0, we are done. However at least one of these must occur, for if u E L1 and P E L are as in (1) and (2) resp. then u y = 0, implying ul yl = - a2 ' y2 nm+l ym+l < 0. Clearly there is nothing special about the (m + 1)-th coordinate

in Km+l and we can substitute m + 1 by any i E {1,...,m+1} in Theorem 4.6. This proves

Corollary 4.7 If L < I-0,ym+1>0 or

b') 3zEK",zT(A,-b)>0,zT(-b)>0 but not both. Clearly, in a') we may require y,n+l = 1 instead of y,n+l > 0. Thus, a') is equivalent to

a") 3xEK",Ax=b,x>0, which is the same as Theorem 4.9 a). Replacing z by -z, we see that b') is equivalent to Theorem 4.9 b). Thus we have shown that Theorem 4.6 implies Theorem 4.9. To prove the converse, let L < K'+1 be an arbitrary subspace of

Km+1. Clearly, L = kerB for some matrix B E E" ('+') and every such matrix can be written in the form (A, -b) with A E K"Xm , b E K". Now it is imediate that Theorem 4.9 implies Theorem 4.6. 0

Given a subset M of K" and a vector b E K", we say that b can be separated from M if there exists a closed linear halfspace H+ of K" such that M C H+ but b f H+. Using this terminology, Theorem 4.9 can be stated as follows: either

a) bEconeA or

b) b can be separated from A

but not both.

53

4.5. DEHOMOGENIZATION

By definition, b cannot be separated from A if and only if it is contained in every closed halfspace containing A, i.e. if and only if it is contained in the intersection of all the closed halfspaces containing A. Thus we get as a corollary from 4.9:

Corollary 4.10 Let A C 1K" be a finite set. Then cone A is the intersection of the closed halfspaces containing A.

Actually, a more general result (which we will not prove) states that every closed cone in K" is representable as an intersection of halfspaces. We shall see later, that if A is finite, then cone A is even representable as the intersection of a finite number of halfspaces hence, that coneA is a polyhedral cone.

Dehomogenization

4.5

To end with our presentation of the Farkas Lemma, we "dehomogenize" Theorem 4.9. Thus, coming back to general polyhedra, which have been our starting point, we will prove that the convex hull of a finite set A C K" is representable as the intersection of affine halfspaces.

Theorem 4.11 Let V, E C 1K" be finite sets and b E 1K". Then either

a) bEconvV+coneE or

b) 3(ro) E1K"}1 s.t. cvO but not both. Here, of course, X _> 0 shall mean that X has a "+"- or a "0"entry in every coordinate. For notational convenience, sign vectors will usually be indexed by elements of a finite set E, and they will always be written as row vectors.

Definition 5.1 Let E 0 be finite. An element X E {+, -, 0}E is called a sign vector (on E). The set of all sign vectors on E is denoted by 2±E

:_ {+, -, 0}E.

The symbols "0", "+" and "-" are referred to as signs. We say that "+" is the opposite sign of "-" and vice versa. For a sign vector X E 2±E we introduce its

positive part X+ {e E E Xe _ +}, negative part X{e E E IXe _ -}, zero part X° {e E E IXe = 0}, and its support suppX := X+ U XWe denote by -X the sign vector which is opposite to X, i.e. whose negative (positive) part equals the positive (negative) part of X. Clearly this notion also applies to sets of sign vectors, thus for S C_ 2±E, -S denotes the set of opposite sign vectors. If X has no negative components on A C E (i.e. x- fl A = 0), we say that X is

nonnegative on A, denoted by XA > 0. XA < 0 and XA = 0 are defined similarily. We will usually write X _> 0 and X < 0 instead of XE > 0 and XE < 0, resp. If S and S' are two sets of sign vectors, then the pair (S, S') is said to have the FARKAS Property, if

(FP) Ve E E either

a) 3XES,eEsuppX andX>0

5.1. SIGN VECTORS

59

or

b) 3YES',eEsuppYandY>0, but not both.

Example 5.2 Let a : 1KE -, 2±E denote the "forgetting magnitudesmap", which associates to every x E 1KE the corresponding sign vector

X = a(x), and let S = a(L) and S'= a(Ll) for some pair of orthogonal subspaces. Then (S, S') has the FARKAS Property (cf. Corollary 4.7). For example, let

A:=

0

1

1

0

1 1

-1 -2

and let L := im A. Let x E IK3 and y = Ax E L. Then, in a sense, the sign vector Y = a(y) reflects the position of x with respect to the hyperplanes (lines) whose normal vectors are the rows As.. In fact, Y, > 0 (< 0), according to whether A. and x are on the same (opposite) side of the line l; :_ {z E 10 A;.z = 0} (cf. Figure 5.1).

12

Figure 5.1

If x E 1K2 is as indicated in Figure 5.1, then the sign vector Y corresponding to y = Ax is Y = (+, -, +, +). Moving the vector x in the plane JA-, we get a complete list of all sign vectors in S := a(L).

CHAPTER 5. ORIENTED MATROIDS

60

There are precisely 17 of them:

X0 = (0,0,0,0) Xl = (+, 0, +, +)

X5 = (+,+,0,-)

X3 = (+, +, +, 0)

Xs = (-, +, -, -)

X2 = (+,+,+,+)

Xe = (+, +, -, -)

X7 = (0,+,-,-)

X4 = (+, +, +, -) and their opposites.

Similarily, one may determine S' = a(L1). Note that,

e.g.

(1, -1, 1, 0)T and (1, 0, -1, 1)T form a basis of L1, hence 1

1

Ll=im -11 -10 0

1

The sign vectors in S' are

Yo = (0,0,0,0)

Y5 = (+,-,0,+)

Yz = (+, -, +, -)

Y7 = (+, 0, -, +) Y8 = +)

Y1 = (0,-,+,-)

Y3 = (+, -, +, 0) Y4 = (+, -, +, +)

Yo = (+,-,-,+)

and their opposites.

Example 5.3 Let G = (V, E) be a digraph and let S and S' denote the sets of sign vectors corresponding to incidence vectors of circuits and cocircuits, resp. Then (S, S') has the FARKAS Property. This is immediate from Theorem 2.12 and Proposition 2.17.

Example 5.4 Let (S, S') and (T, T') have the FARKAS Property and

assume that S C_ T and S' C T. Then every pair (7Z, R') "in between", i.e. such that S C_ R C T and S' C VC T', has the FARKAS Property.

Finally, let us note, that there are really trivial examples, too. Let S consist of a single sign vector Y = (...... +), and let S' = 0. Then, obviously, (S, S') has the FARKAS Property. The existence of such degenerate examples indicates that we should impose further conditions, in order to exhibit interesting structures.

5.2. MINORS

5.2

61

Minors

As indicated in the last paragraph, simply requiring the FARKAS Prop-

erty to hold for a pair (S, S') does not lead to an interesting theory. So what other "structural" properties of vector spaces can be translated to sign vectors? One of the most important characteristics of vector spaces is that the intersection of two (or more) vector spaces is a vector space again. In particular, if L < KE is a subspace, and e E E, then

{xELI x.=O} is a subspace again. If L = ker A, for some matrix A whose columns are indexed by E, this operation corresponds to deleting column e from A. Thus let A \ e denote the matrix obtained from A by deleting column e. Then

L \ e : = ker(A\e) O,UG0 but not both. By the definition of minors, this means that either

(a) 3UEO.F, eEsuppU, URUG>0, UB=*, UW=O or

(b) 3YEO.F, eEsuppY,YRUG>-O,YB=O,YW=* but not both. This is equivalent to the condition in the claim.

Remark 5.13 MINTY's Lemma may be visualized as indicated in the note following Definition 5.6: If E = R U G U B U W is a partition of

E and, say, e E R, then either R

G

B

W

e

3X=+®...®e...e *...* 0...0 E-F or

3Y=+®...®e...e 0...0 *...*

E F'

but not both.

Here, clearly, "®" stand for "+" or "0", "e" stands for "-" or "0" and "*" stands for "+" or "-" or "oil. At the first glance, Proposition 5.12 seems quite "technical". (Indeed, it will turn out to be a powerful technical tool.) We will see later how MINTY's Lemma can be used, to prove the "Max flow- min cut"Theorem. Then, each of the four colours will have a special meaning and the theorem may be understood more clearly.


66

5.4

Abstract Orthogonality

Recall that our original problem (cf. the introduction to Chapter 5) has been the following: Given an OM (F, F'), does there exist a corresponding pair of orthogonal subspaces (L, L1)? Here we will give an answer to a related (but simpler) question, namely: Given an OM (F, F'), does there exist a kind of orthogonality relation between

F and F'? If x, y E 1 E, then x and y are orthogonal, if their inner product xT y equals zero. Since there is no way to "add" signs, there seems to be no reasonable way to mimic this definition for sign vectors. So we have to try to get on with a weaker one. Note that if x, y E KE "agree in sign" in some coordinate e E E (e.g. x,. > 0 and y. > 0), then x 1 y implies that x and y "differ in sign" in some other coordinate f E E. This motivates the following definition:

Definition 5.14 Two sign vectors X, Y E 2±E are called orthogo-

nal, if

(X+nY+)u(X n Y):A 0 .(X+nY-)u(X nY+)#0. This is denoted by X I Y. If S C_ 2±E, then X E 2±E is called orthogonal to S, denoted by X 1 S, provided X 1 Y for every Y E S. The set

S1= {XE2±EIXIS} is called the orthogonal complement of S. If S' C S1, then we say that S and S' are orthogonal.

It is an easy exercise to show that S C S11 := (S1)1 for every set of sign vectors. Moreover, the map S -> S1 is inclusion reversing, i.e. S C T implies S1 D T1.

Example 5.15 Let C and C* denote the sets of incidence vectors of circuits and cocircuits, resp., of a digraph G = (V, E). Let x E C and y E C*, and let X and Y denote the corresponding sign vectors. Then X 1 Y expresses the fact that if the circuit x passes the cut y in one direction, it must also pass it in the opposite direction.

Example 5.16 Let L < HE be a subspace. Since, as we noted already, "abstract orthogonality" between sign vectors is weaker than "ordinary orthogonality" between vectors, it follows that

a(L)1 J o (L1).

5.5. ABSTRACT ELIMINATION PROPERTY

67

We will show later that actually equality holds, indicating that the abstract version of orthogonality is (still) quite reasonable. The reader is invited to check this in the special case of Example 5.2. Here we will content ourselves by showing that orthogonality between the is an easy consequence of two "parts" F and.F' of an OM MINTY's Lemma:

Theorem 5.17 Let.F, .F' C 21E be two sets of sign vectors. Then .F and F' are orthogonal if and only if for every partition E = R U G U B U W and every e E R U G at most one of the alternatives (a) and (b) in (MP) holds.

Proof. Let E= R U G U B U W be a partition of E, and let e E E. If there exists X E F and Y E F' as in a) and b) of Proposition 5.12, then e E (X+nY+) U (X- nY-), while (X+nY-) u (X- nY+) = 0. Thus F and .F' are not orthogonal. Conversely, suppose that F and .F' are not orthogonal. Let X E.F and Y E.F' such that X and Y are not orthogonal. Assume that, say, (X+ n Y+) u (X- n Y-) i4 0 and

(X+ n Y-) u (X- n Y+) = 0. Now let R:=X+nY+, G:=X-nY-, B := Y°, W := X° \ Y° and e E RUG. Then condition (a) of Proposition 5.12 is satisfied as well as (b). We obtain at once two corollaries:

Corollary 5.18 If i. e..F' C Fl.

is an OM, then.F and.F' are orthogonal,

Corollary 5.19 If (.F, .F') is an OM, so is

Proof. If (.F, .F') is an OM, then F' C_ Fl. Since MINTY Property, F' C Fl implies that for the pair

has the at least one of the alternatives (a) and (b) in (MP) holds. On the other hand, Theorem 5.17 shows that at most one of them can hold.

5.5

Abstract Elimination Property

In linear algebra, one usually considers single linear vector spaces rather than pairs of orthogonal vector spaces (L, L1). There one starts with a set L of "vectors" (and some field F) and defines L to be a vector space, if certain axioms are satisfied. In the sequel we will try to proceed in a similar way: We will start with a set of sign vectors and define it to be an "oriented matroid", if it has certain properties.


68

Since the essential axiom for vector spaces is, that they are closed with respect to addition of vectors - an operation which doesn't make sense for sign vectors -, we have to look for another property of vector spaces which is more "combinatorial" in nature, allowing to interprete it in the context of sign vectors. This property will get the name "elimination property". Again, let us first consider some examples in order to see what this property is like.

Example 5.20 Let L < IKE be a subspace, and let x, y E L. We say that e E E separates x and y, if x and y have opposite signs in coordinate e. Equivalently, x and y lie on opposite sides of the hyperplane He :_ {w E 1K' I w, = 0}, i.e. HH separates x and y. If e separates x and y, then the line segment [x, y] intersects HH in a point (Vector) z = Ax + µy, A > 0, u > 0 and A + µ = 1. The e-th coordinate of z is equal to zero, and therefore, we will say that z arises by eliminating e between x and y . If in addition x f # 0 for some

f E E which does not separate x and y, then (since A > 0 and p > 0) the f -th coordinate of z has the same sign as the f -th coordinate of x. We therefore say, that z arises by eliminating e between x and

y, fixing f. . Example 5.21 Let G = (V, E) be a digraph, let L denote its circuit space, and let 2 denote the corresponding set of sign vectors. ell

eio

e6

e4

es

Figure 5.2 If G is as indicated above with edgeset E = {ei, ... , ell}, then, e.g. the sequence o f edges (ell e2, ... , e8) forms a closed path and hence the

incidence vector x E IKE of this path is an element of the circuit space of G. Similarily, the incidence vector y of the circuit (e2, e9, e67 elo, ell)

is an element of L. The corresponding sign vectors X = v(x), Y = o,(y) in.F are, resp.:

X = (+,+,+,-,+,+, +, 0, 0, 0) Y = (0,+,0,o,0,-,0,0,+,-,+)

5.5. ABSTRACT ELIMINATION PROPERTY

69

Obviously, X and Y are separated by e6 and, e.g. el is an element of suppX which does not separate X and Y. Then for example, the sign vectors

Z = (+,+,+, 0, 0, 0, 0,+, 0, 0, 0) E.F and Z' _ (+,+, 0, 0, 0, 0,-,+,+, 0, 0) EF corresponding to the circuits (el, e2, e3, eg) and (el, e2, e9, e7, es) are obtained by eliminating e6 between X and Y, fixing el.

Definition 5.22 Let X, Y E

21E. Then e E E separates X and Y if X. and YY are opposite signs. The set of elements e E E separating X and Y is denoted by sep(X, Y). Suppose that e, f E supp X U supp Y such that e separates X and Y, but f does not. Then Z E 21E is said

to be obtained by eliminating e between X and Y, fixing f, if

eVsuppZ, f EsuppZ, Z+CX+UY+ andZ- CX- UY-. The set S C 21E is said to have the elimination property, if the following holds:

(EP) If X, Y E S, e, f E supp XUsuppY such that e separates X and Y, but f does not, then there exists a Z E S which is obtained by eliminating e between X and Y, fixing f. (Note that if X and Y axe separated by some e E E, then there always

exists a nonseparating element f, unless X = -Y.)

Theorem 5.23 Let (F,.P) be an OM on E. Then .F has the elimination property.

Proof. Let X, Y E F and suppose that e and f are as above. To apply MINTY's Lemma, let R G

B W

(X+ U Y+) \ (X U Y-), (X U Y) \ (X+ U Y+), [(X+ n Y) u (X n Y+)] \ e,

X°nY°Ue.

Thus, in particular, f E R U G, and hence either

a) 2ZE.F, f EsuppZ, ZR>0, ZG 1, we will show that we can write Y as Y = Yi o...oY' f o r some 1 1 , . .. , Y' E F such that the number of elements separating some of the sign vectors i i , . .. , Y' is strictly less than t. Thus, let e E E separate Y, and Y, i < j. For every f E TO fl suppY, let

76


Y(f) E F be obtained by eliminating e between Y and Y fixing f . Then we replace Y in Yl o... o Yk by the composition of all these Y(f)'s to get Y = Yl o ... o Y,,. It is easy to see that the number of elements separating any two of Yi..... Y is < t. Thus, if we proceed eliminating e between every pair Y, and Y such that e separates Y, and Y, we will finally end up with a representation Y = Y, o ... o Y' such that e does no longer separate any two of Yl..... Y'. 0

There are many consequences of the Composition Theorem. We just list some of them below:

Corollary 5.37 Let (F, F') be an OM on E. Then (elemF)1 = .F1 Proof. F -L C (elemF)1 is trivial. To prove the converse inclusion, let U E (elemF)1, i.e. U 1 X for every X E elem.'. It is easily seen that this implies U 1 Y for every conformal sum Y = Xl o ... o Xk of elementary vectors X; E elem.. Hence U E .F1. 0

be an OM on some set E. Then Corollary 5.38 Let (.F, elem.), and hence (elem.F,F') and (elem., elem.') are OMs, too.

Proof. We will verify MINTY's Property for (.F, elem.') (cf. Propo-

sition 5.12). Let E = R U G U B U W be a partition of E, and let e E R U G. Consider the two alternatives

a) 3UEeEsuppU,UR>0,UG0,ZG0,YG (0,Yw=*. Thus, for every e E supp X 3 Z(e) E elem7-1, e E supp Z(e), Z(e)R > 0, Z(e)G:5 0, Z(e)w = 0.

The composition of all these Z(e)'s, e E supp X, is equal to X. Hence,

X E span(elem?-1) = spanf. Since X has been an arbitrary element of elem.', this shows that elem.' C span W, and consequently,

using the composition theorem, span. = span(elem.F) C span H. By symmetry, span. = spanl-1. This in turn gives elem.' _ elem(span.'F) = elem(spanl-1) = elem11.

Corollary 5.41 Let

be an OM. Then span.= .1:'1.

Proof. By Corollary 5.19, (F,.F1) is an OM. Furthermore, we con-

clude from Proposition 5.40, that span.' = span F1. But F1 is obviously closed under conformal sums, i.e. span F1 = F1 0

Corollary 5.42 If L < 1KB is a vector space, then o-(L1) = a(L)1. Proof. Since (v(L), o-(L1)) is an OM, it follows from Corollary 5.41,

that span(u(L1)) = v(L)1. However, since L1 is a vector space, a(L1) is obviously closed under composition, i.e. span(v(L1)) _ 0.(Ll).


78

5.8

Elimination Axioms

Recall that our aim has been to derive an alternative definition of oriented matroids by considering only a single set F C 2±E rather of systems of sign vectors. Thus our goal than a dual pair is to characterize those sets F C 2±E for which there exists a dual Jc' C 21E such that (JI', F') is an OM. The results of the last section imply that it suffices to characterize the sets elem F resp. span F, where (F, F') is an OM. Indeed, this has been the first approach to oriented matroids in the literature. We start with defining oriented matroids in terms of their elementary sign vectors.

Definition 5.43 Let 0 # E be finite and C C 2±E. Then C is called an elementary oriented matroid ("OM"), if the following holds: a) 0 V C and C is symmetric,

b) C = elemC, c) C has the elimination property.

Proposition 5.44 If (.F, P) is an OM, then elem. is an elementary OM. Proof. Corollary 5.39. 0

We refer to previous examples for an interpretation of the elimination Axiom c). In particular, the reader may try to find an "elementary" proof for the fact that the sign vector corresponding to the elementary vectors of some subspace L < KE form an elementary OM. This is not completely trivial (once you agree that "elementary" means "without using the Composition Theorem"). Thus, so far, we have given two definitions of oriented matroids (which are not yet proved to be equivalent), and neither of them is trivially satisfied by vector spaces (resp. their elementary vectors). Our third (and last) definition of oriented matroids, which we are going to introduce below, does have this property. In fact, the socalled "complete" OMs, which will be defined in a minute, are such that one can recognize subspaces of KE as oriented matroids at once.

5.9

Approximation Axioms

Let L < KE be a subspace. In Example 5.20, we have shown that u (L) has the elimination property. There we considered x, y E L such

5.9. APPROXIMATION AXIOMS

79

that x and y were separated by an element e E E and we observed, that we can eliminate e between x and y by fixing some f E supp x U suppY which does not separate x and y, thus obtaining the vector

z = Ax + µy, A and p > 0, with z. = 0 and zf having the sign of x f + y f. Actually, these arguments prove more than the simple fact that a(L) has the elimination property. Since the element f E E is arbitrary, they prove that z f has the same sign as x f + y f for every f E E which does not separate x and y. In terms of sign vectors, this can be stated as follows:

Proposition 5.45 Let L < IKE be a subspace, and let 0 := o ,(L). Then, given X, Y E 0 and an element e E sep(X, Y), there exists a sign vector Z E 0 such that Ze = 0 and Zf = (X o Y) f for every f E E \ sep(X, Y). 0

Example 5.46 Let L := im A for some matrix A E KE"" and 0 := o(L). Furthermore, let u, v E IK", and let X := a(Au), Y := a(Av). If e E sep(X, Y), this means that u and v lie on opposite sides of the Nowletw:=H.fl[u,v]. hyperplaneHH:={wEIK" I Then, obviously, Z = a(Aw) is as above. One might say, that we obtained w by approximating u and v on He. This terminology can be justified by observing that w minimizes the term IIw'-u11+11w'-vII,

w'EHH.

Definition 5.47 Let X, Y E 2±E be two sign vectors, and let e E sep(X, Y). Then Z E 2±E is said to be obtained by approximating X

andY on e, if Z. = 0 and Zf = (XoY)f for every f E E\sep(X,Y). A set S C 2±E is said to have the approximation property, if the following holds:

(AP) Given X, Y E S and e E sep(X, Y), there exists Z E S which is obtained by approximating X and Y on e. Note, that (AP) is essentially some kind of "strong" elimination property. If Z is obtained by approximating X and Y on e, this means

that Z is obtained by "eliminating e between X and Y, fixing all elements f which do not separate X and Y". More precisely, the connection between the elimination and the approximation property is the following:

Lemma 5.48 If S C 211 is symmetric and has the elimination property, then span S has the approximation property.


80

Proof. Let S have the elimination property, and let X, Y E span S be separated by some e E E. Let us first assume that X, Y E S. Then, by assumption, for every f E supp X UsuppY which does not separate

X and Y, there exists a Z(f) E S which is obtained by eliminating e between X and Y, fixing f. The composition Z of all these sign vectors Z(f) is an element of span S and it obviously approximates X and Y on e. Next consider the general case X, Y E span S. Since S has the elimination property, Theorem 5.36 implies that X and Y can be

written as conformalsums X=X1o...0Xk and Y=Ylo...oY of vectors in S. We proceed by induction on k + 1. If k = 1 and l = 1, i.e. X, Y E S, the argument from above applies. Thus assume that, say, k > 1. Let X':= X1 and X" := X2 o ... o Xk. By induction, there exist Z', Z" E span S which are obtained by approximating X' and Y, resp. X" and Y on e. Then it is easy to see that Z := Z' o Z" is an element of span S which is obtained by approximating X and Y on e. 0

We now give the definition of oriented matroids by means of approximation axioms. As we will see, the definition below characterizes

the sets span.F, where (F, F') is an OM.

Definition 5.49 Let O C 2±E. Then O is called a complete oriented matroid ("OM") if the following holds: a) 0 E 0, and 0 is symmetric,

b) O=span O, c) O has the approximation property.

Recall that if (F, .P) is an OM, then (span., F') is also an OM and span. = (F')1 by Corollary 5.41, hence span F is maximal with this property. This justifies the notion "complete". Theorem 5.50 If C is an elementary OM, then span C is a complete OM. Proof. Follows from Lemma 5.48. 0

Thus in order to prove that all three definitions of oriented matroids are equivalent, we are left to verify that a complete oriented matroid 0 gives rise to an OM. Essentially, this means to show that the pair (0, Ol) has the FARKAS Property (FP). Thus we are left to give a proof of FARKAS' Lemma in the abstract setting of sign vectors.

5.10. PROOF OF FARKAS' LEMMA IN OMS

5.10

81

Proof of FARKAS' Lemma in OMs

We are going to show that, given a complete oriented matroid 0, the pair (O, O1) is an OM. From Definition 5.8 of OMs, it is clear that

we have to deal with minors of (O, O1). Recall that we defined a minor of a pair (S, S') to be a pair (S \ I/J, S'\ J/I). In other words, we defined the contraction to be the operation "dual" to deletion and vice versa. (Here "dual operation" means "corresponding operation on the orthogonal complement".) This has been justified by the fact that (L \ I/J)1 = L1 \ J/I for every subspace L < KE. However, we did not prove the corresponding result for general sets of sign vectors!

Lemma 5.51 Let 0 C 2±E such that 0 is symmetric and 0 has the approximation property. Then for every e E E we have (0 \ e)1 = O1/e. In particular, this holds, if 0 is a complete OM. Proof. "3" follows immediately from the definition of minors. To prove the converse direction, let U E (0 \ e)1. We want to show that there exists U E O1 such that Uf = Uf for every f E E \ e. Clearly,

there are just three possibilities for U: Either Ue = +, U. _ - or U. = 0. We denote this by U = U + e+, U = U+ e- and U = U + eo, resp. Assume that non of these is an element of O1. Hence there exist X, Y, Z in 0 such that U + e+ x X, U + e- Y and U + eo .x Z. By reversing signs, if necessary, we may assume that the situation is as follows: e

U+e+ = + ... +- ... - 0 ... 0 + U+e- = + ... +- ... - 0 ... 0 U+eo = + 0 ... 0 0 +...ED e...e*...* + X= ED

Y = ® ...®e...e*...*

Z = +ED ...®e...e*...* We may assume w.l.o.g. that Z. = 0. For, if Z. 0, we may approximate Z and X or Z and Y (according to whether ZZ = or ZZ = +) on e in order to get a zero entry in coordinate e of Z. Thus we may in fact assume that Z. = 0. But this immediately gives

a contradiction, since then Z = Z \ e E 0 \ e is not orthogonal to U E (O \ e) 1.

0 Surprisingly easy to prove is the following "dual" result:


82

Lemma 5.52 Let 0 C 21E have the approximation property. Then for every e E E we have (0/e)1 = O1 \ e. In particular, this holds, if 0 is a complete OM. Proof. Again, "p" follows from the definition of minors. To prove "C", let Z E (0/e)1 and define Z E 2±E by Ze := 0 and Zf := Zf for every f E E \ e. We have to show that Z E O1. Thus, let Y E 0, and let Y= Y \ e E O/e. Then Y I Z follows immediately from k I which proves the claim. 0

Lemma 5.53 Let 0 be a complete OM. Then every minor and every reorientation of 0 is a complete OM. Proof. This follows immediately from the definitions. 0

Thus, in order to show that (0, 01) is a an OM, it is sufficient to show that (O, O1) has the FARKAS Property. This can be done in exactly the same way as we proved FARKAS' Lemma in vector spaces:

Lemma 5.54 Let 0 C 2fE such that 0 is symmetric and 0 has the approximation property. Then for every e E E there exist X E 0 and Y E 01 such that X > 0 and Y > 0 and (X o Y)e = +.

Proof. The proof is by induction on n = AEI. The claim is trivially

true for n = 1. Thus let n = JEI > 2 and assume the result holds for n - 1. Let e E E, and choose f # e, f E E. We consider both 0 \ f and 0/f. These are complete OMs by Lemma 5.53. Applying our inductive assumption to 0 \ f , we find (using Lemma 5.51) that

ef 3X = ®0®...® 3Y =

EO

EOlwith (XoY).=+

Similarily, considering O/ f (and using Lemma 5.52), we get

3X' = ®*ED ...®

E0

31" = ®0®...® EOlwith (X'oY')e=+ Since X' I Y, we see that X' and Y cannot both have a "-" entry in coordinate f. The claim follows. 0

Theorem 5.55 (FARKAS' Lemma) Let 0 be a complete OM. Then (0, O1) has the FARKAS' Property. Hence, since complete OMs are closed under taking minors and reorientation, (0, O1) is an OM. 0

5.11. DUALITY

5.11

83

Duality

The equivalence of the three definitions of oriented matroids essentially says, that if F C 2±E has some "nice" properties (such as elimination or approximation property), then the orthogonal complement F1 is nice, too. This principle of "duality" or "orthogonality" is fundamental in the theory of oriented matroids. The duality theorems, listed below, are quite hard to prove from scratch, but they turn out to be immediate corollaries of our previous results.

Theorem 5.56 If 0 is a complete OM, then so is 0'. Furthermore, 011 = 0. (01 is called the dual of 0.) Proof. If 0 is a complete OM, then (0, 01) is an OM by Theorem 5.55. Therefore, elem O1 is an elementary OM and span(elem 01) is a complete OM. Since (01, 0) is an OM, (elem(01), 0) is an OM, too, by Corollary 5.38. Thus, Corollary 5.41 states that O1 equals span(elem 01), and hence O1 is a complete OM. Furthermore,

0 = span 0 = 011 by Corollary 5.41. 11

Theorem 5.57 Let C be an elementary OM. Then C1 is a complete OM and C* := elemC' is an elementary OM. Moreover, C** = C.

Proof. Let C be an elementary OM. Then span C and (by Theorem 5.56) (span C)1 are complete OMs. Since (span C)1 = C1 (cf. Lemma 5.33), this proves the first part of the claim. Moreover, C** _ elem((elem(C1))1) = elem(C11) = elem(spanC) = elemC = C. In Chapter 7, where we shall see how to formulate "Linear programs" in OMs, duality will be studied more extensively. In most cases, we will deal with complete OMs only. So, for simplicity, let us make the convention that the term "OM" shall apply to all three kinds of oriented matroids. It will always be clear from the context or from notation (C or 0, resp F) whether we consider elementary or complete OMs.

5.12

Further Reading

It is always difficult to say who has been the first. Among those who prepared the way to oriented matroids, one should mention


84

R.T. ROCKAFELLAR The Elementary Vectors of a Subspace of IR", in Combinatorial Mathematics and its Applications, pp. 104-127 in "Proc. of the Chapel Hill Conference", Chapel Hill (R.C. Bose and T.A. Dowling, eds.) University of North Carolina Press, Chapel Hill (1969).

G.J. MINTY On the Abstract Foundations of the Theory of Directed Linear Graphs, Electrical Networks and Network Programming, Journal of Mathematical Mech. 15 (1966), pp. 485-520.

Rockafellar, in the above mentioned article, seems to have foreseen much of the developement in oriented matroid theory. The paper of Minty presents the concept of digraphoids, an abstraction of duality in directed graphs, as we discussed it in Chapter 2. Oriented matroids, as we know them today, have been introduced independently by R. Bland, M. Las Vergnas, A. Dress, J. Folkman and J. Lawrence, cf. R. G. BLAND A Combinatorial Abstraction of Linear Programming, Journal of Combinatorial Theory, Series bf B 23 (1977), pp. 33-57.

R. BLAND, M. LAS VERGNAS Orientability of Matroids, Journal of Combinatorial Theory, Series B 24 (1978), pp. 94-123. J. FOLKMAN, J. LAWRENCE Oriented Matroids, Journal of Combinatorial Theory, Series B 25 (1978), pp. 199-236.

A. DRESS Chirotopes and Oriented Matroids, Bayreuther Mathematische Schriften 21 (1986), pp. 14-68.

The approach taken by Bland and Las Vergnas was to develop oriented

matroids out of what is called "matroid theory", a theory that had been well established already at that time. Informally, matroid theory investigates systems of "sign vectors" having coordinates equal to "0" and "1" (representing a " 0") only. Matroids arise as an abstraction of linear subspaces of K' in the very same way as oriented matroids

do if one does not only apply the "forgetting magnitudes map" a : K" - {0, -}-, -}n, but rather a "forgetting magnitudes and signs map"

Q : K" --p {0,1}". Thus, if one replaces the signs "+" and "-" by "1" in the definition of matroids, one arrives at a definition of matroids. For example, the elimination axiom yields the following characterization of matroids: A matroid is a system C C {0,1}E such that

(i) 0 (ii) C = el em C


85

(iii) If X, Y E C and Xe = Y, = 1, X f = 1, Yf = 0, then there exists Z E C which is obtained by eliminating e between

X and Y, fixing f, i.e. ZZ = 0, Zf = 1 and suppZ C supp X U supp Y.

The beginning of matroid theory dates back to the early work of Whitney, cf. H. WHITNEY The Abstract Properties of Linear Dependence, American Journal of Mathematics 57 (1935), pp. 509-533. Since 1960, the field has developed rapidly and has become one of the major research areas within combinatorial optimization. From the above definition of matroids it is obvious that every ori-

ented matroid has an underlying ("unoriented") matroid, that is obtained by "forgetting signs". Conversely, given an arbitrary matroid, one may try to "orient" it, i.e. replace every "1" entry in each vector

by a "+" or "-" coordinate such that the resulting system of sign vectors is an oriented matroid. This is not always possible, i.e. there do exist "nonorientable" matroids. Nonetheless, it is clear that it is easier to develop oriented matroids out of matroid theory than starting from scratch (i.e. from linear subspaces of K"). We did not want to introduce matroid theory, however, since this is somewhat aside the other topics treated in this book, such as linear programming and polyhedral theory. Therefore, we preferred to choose the direct way to oriented matroids, which is abstract linear duality. This approach, which is the most convenient one for those not familiar with matroid theory, has been worked out by several authors, cf. [28], [70], [78]. Andreas Dress [60] developed in 1977 from a definition in terms of

the Grassmann-Plucker-relations a new structure which he together with the organic chemist Andree Dreiding and his mathematical col-

laborator Hans Haegi used to investigate the chirality of organic molecules. Later on it turned out that these chirotopes are equivalent to oriented matroids. Moreover, Lawrence [121] showed that Gutierrez-Novoa's (cf. [98]) n-ordered sets are another variant of this approach. They derive oriented matroids by introducing an orientation on the bases of the underlying (unoriented) matroid. For those who are familiar with matroid theory, let us briefly outline the main idea of this approach. Let E be a finite set, d < A EI and let det : Ed -+ {0, ±1} be a nonvanishing alternating map. Then (E, det) is called a "base oriented" matroid, if for all (fi, ... , fd) E Ed and (g1, ... , gd) E Ed the following implication ("Grassmann-Plucker relation") holds:


86

(i) Vj: det(gJ, f2, ... , fd) det(gl,... , 9i-1, fi, 9i+1, ... , 9d) > 0 imp lies

(ii)

det(fi, ... , fd) - det(9i, ... , 9d) > 0

.

It is easy to see that the sets {b1, ... , bd} for which det(bi, ... , bd) A 0 form a set 13 of bases of some matroid. Indeed, let B = {b1, ... , bd} E 13 and B' _ {b'1, ... , b'd} E B and, say, bl E B \ B'. Since

det(bl, ... , bd) det(bl, ... , Yd) # 0

,

the above condition implies that for some j we have det(b', bs, ... , bd)

:0

.

Hence, B \ bl U b' E 8. This shows that 13 is indeed the family of bases

of a matroid M of rank d. In the following we will briefly sketch how one can construct from a given base oriented matroid (E, det) the set of sign vectors of the corresponding elementary oriented matroid C and its dual C*. Details can be found in Lawrence [121]. Given a circuit C of M, let B = {b1,. .. , bd} be any basis such that C is a fundamental circuit with respect to B, i.e. C C B U bo for some element bo E E. Define

X+ := bo U {bi I det(bl.... , bi-i, bo) bi+1, ... , bd) = -det(big .... bd)}

X-

._ {bi I det(bl.... , bi_1, bo, bi+1, ... , bd)

=

det(bl,... , bd)}

Then X+IJX- is a partition of C. Using the Grassman-Plucker relations, one can show that this partition is independent from the particular choice of the basis B. The set C of sign vectors X defined this way is an elementary OM whose underlying matroid is M. The dual C* can be obtained from (E, det) as follows. Let C* be a cocircuit of M and let F = E \ C* be the corresponding hyperplane. Let {f2,..., fd} be a basis of F. Then Y+

{e det(e, f2, ... , fd) _ +1}

Y-

{e det(e,f2,...,fd)_-1}

and

form a partition of C*. Again, one can show that this does not depend

on the particular choice of fl,... , fd_1. The set C* of sign vectors Y obtained this way is the elementary OM dual to C. Note that C-LC'. In fact, let X E C and Y E C* as above and assume that, say, bo E X+ n Y+. Thus det(bo, fx, ... , fd)

_ +1 .


87

Assume w.l.o.g. that, say, det(bl,... , bd) = +1. Hence, det(bo, f2, ... , fd)

det(bl,... , bd) = +1

.

The Grassman-Plucker relations now imply that for some i, 1 0}, define the same optimal value

z*. - Well, not quite.... Recall that we have shown this under the additional assumption, that one of them (in fact, the primal one) is feasible. However, it may happen that both programs are infeasible, in which case their optimal values are -oo and +oo, resp..

92

CHAPTER 6. LINEAR PROGRAMMING DUALITY

Example 6.2 Let A = (_1 -1), b = (=i) and c = (01). Then both max{cx I Ax < b} and min{bTU I ATU = c, u < 0} are easily seen to be infeasible. As we remarked earlier, the symmetry between the relations (6.2) and (6.3) suggests, that duality between LPs is a symmetric relation, too. We can not prove this, however, before having defined the "dual" of a minimization problem. Thus consider the LP min{cx I Ax _< b}. What should be a natural definition for its dual? Of course, we could go along the same way as above, working through all the PZ- and Pz reformulations. The reader is invited to do so - or just to believe that the following simple reasoning arrives at the same end: Note that min{cTx I Ax < b}

z'

- max{-cTxI Ax0} max{bTU I ATU = C, U < O}.

Definition 6.3 The dual of the LP min{cTxI Ax 0}, i.e. the dual of (6.1) and rewrite it as c

AT

min{bTu

-c

-AT Iu < k-I

0

where I denotes the unit matrix. The dual of this is

fx+ max { (cT, -cT, 0) j x-

kx°

x+

(A, -A, -I)

x-

= b and x+, x-, x°
0 whenever xi = 0. Thus, up to scaling by a factor 1/z;,+1, z' is improving with respect to x. We have shown that x is optimal if and only if there is no z E L which is improving with respect to x. This notion readily carries over to general oriented matroid programs: A "feasible" solution X E 0 will be called "optimal" if there is no Z E 0 which is "improving" with respect to X.

6.2. THE COMBINATORIAL PROBLEM

99

Similarily, the notion of unboundedness can be defined for general

oriented matroid programs. Suppose there exists a z E L such that zo = 0, zn+1 = 1 and z; > 0 for all i = 1, ... , n (hence z is improving with respect to every feasible x). Then, if (6.11) is feasible, say, x E L

is a feasible solution, we see that x + ez is feasible for every e > 0. Thus, in particular, (6.11) is unbounded. We are now facing a long and boring definition.

Definition 6.7 Let O be an OM on E = {0, ... , n+1}. Then X E 0 is called a (primal) feasible solution, if X0 = + and X1,.. . , X > 0. Y E O1 is called a (dual) feasible solution, i f Y1, ... , Y < 0 and Furthermore, O(O1) is called feasible, if there exists a feasible solution X E 0 (Y E O1), otherwise it is called infeasible. Given a feasible solution X E 0, we say that Z E 0 is improving(with respect to X), if Zo = 0, and Z, > 0, whenever X; _

(i = 1,.. . , n). Similarly, given a feasible Y E O1, we say that Z E O1 is improving (with respect to Y), if Zo = -, Zn+1 = 0 and Z; < 0 whenever Y, = 0 (i = 1,..., n). A feasible solution X E 0 (Y E 0

O1) is optimal, if there is no Z E 0 (Z E O1) that is improving with respect to X (Y). If X and Y are feasible solutions of 0 and O1, resp., then X and Y are called complementary, provided suppX fl

suppY C {0, n + 1}. Finally, 0 is called unbounded provided it is feasible and there exists a Z E 0 such that Zo = 0, Zn+1 = + and Z, > 0 for all i = 1, ... , n. 01 is called unbounded, if it is feasible and there exists Z E O1 such that Zo = -, 0 and Z; < 0 for all i = 1, ... , n. 0 (01) is called bounded, provided it is feasible and not unbounded.

A dual pair of oriented matroid LPs will be denoted by max Xn+1

XEO Xo=+,X1i...,Xn>0

min Yo and

YEO1 Y1,...,Yn 0 for all i = 1,. .. , n. Let x and z be corresponding elements

in L. Then x + ez is feasible for all c > 0, showing, that (6.11) is unbounded. The converse is, in fact, nontrivial: Suppose that (6.11) is unbounded. Then, by Theorem 6.4, (6.12) is infeasible. Thus, 0' is infeasible, i.e. ,A

YEO'

Applying MINTY's Lemma, this yields

3Z=00...®+, ZEO, as required. 0

Let us summarize, what we have done so far. We started out with the general "ordinary" LP (6.1) and transformed it, together with its dual until we arrived at (6.11) and (6.12). All of these single transformation steps were such that an obvious correspondence between the optimal (non optimal) solutions of the various reformulations was maintained. Thus, in particular, (6.1) is feasible, unbounded, etc., has or does not have optimal solutions if and only if (6.11) does. Proposition 6.8 says, that a similar relationship holds between (6.11),(6.12) and the corresponding oriented matroid LPs. This time, however, we have not just reformulated the problem, but we have also reduced the set of feasible solutions to a finite set! Therefore, the reader may wonder about whether our original optimization problem has not become trivial in a certain sense. For example, suppose we are given a feasible solution x of (6.11) and we want to decide whether it is optimal or

not. All we have to do is, to look at 0 = a(L) and check whether among the (finitely many!) elements of 0 there is one which is improving with respect to X = a (x). Of course, this would be an easy


101

thing, once we are given 0 explicitely, i.e. a list of all its sign vectors. However, in general, we are not (in fact, deciding whether a given sign vector Z E 21E is an element of 0 = a(L), turns out to be as difficult as solving "ordinary" LPs.1) Thus, oriented matroids do not seem to

be useful for solving LPs. They are, however, quite useful in that they provide us a means for proving LP-duality theorems in a very clear and systematic way, revealing precisely those ("combinatorial") properties of K" that make the whole theory work. Let us start our investigation of oriented matroid LPs by introducing some further notation. Sometimes it will be convenient to consider such Linear Programs for minors of 0 as well. If, e.g. I = {1, ... , k}, J = {k + 1, ... 11, then the LP associated to O \ I/J and its dual will be denoted by J

I

X = +0 0 *®...® }E0 resp.

Y = j*...*0...0e...e-E01. Furthermore, one may also consider oriented matroid LPs associated

to reorientations of 0 or a minor of O. For example, if I, J, K are disjoint subsets of {1, ... , n} then the LP corresponding to gO \ I/J will be denoted by I

J

K

X = +0

T EB...®? EO

and its dual is

Y = j *... * 0...0®...®e...e-EO1. Of course, we may also reorient coordinates 0 and n+1. For example, if in the last problem above coordinate n+1 is reoriented, then this turns the maximization problem into a minimization problem. Therefore,

the LP that is obtained from the above by reorienting coordinate n +1 will be denoted by 'Indeed, Z E 21' is an element of 0 = a (L), if and only if each of the following LPs

xj>0 x,0

x, 0, if i E Z+ and xi < 0, if i E Z-. In this case, if z denotes the sum over all these optimal solutions,

Z = o(z) E 0.

CHAPTER 6. LINEAR PROGRAMMING DUALITY

102

I

X = -} 0

K

J

* e e®...®j ECM

and its dual will then be

Y = j *... * 0...0®...®e...e+EOl. Finally, we will sometimes allow other coordinates to play the role of coordinates 0 and n + 1. This probably needs no further explanation. As an example consider the following LP

x = 0...0*...* T e...e+®...®E O and its dual

Y = *...*0...0-®...® I e...eEO1. The following lemma provides some simple facts about oriented matroid programs:

Lemma 6.9 Let O, O j- be a dual pair of OMs on E = 10,...,n+111 and consider the two dual oriented matroid LPs

X=+$...®I E0

and

Y=1e...e-EOl.

Then the following holds:

(i) A feasible solution X E 0 is optimal if and only if there exists a feasible solution Y E Ol such that X and Y are complementary. Dually, a feasible solution Y E Ol is optimal if and only if there exists a feasible complementary X E O.

(ii) If X E 0 is feasible, and X E 0 is optimal, then X,+1. (Here ">" denotes the obvious ordering on {+, -, 01,

i.e. - < 0 < +.) In particular, if X and X' are optimal solutions, then X,,,+1. Again, the dual statement, i.e. the corresponding statement about feasible and optimal solutions Y E Ol is also true.

(iii) If X E 0 and Y E Ol are primal/dual optimal solutions, then

Yo.

(iv) If 0 is feasible, then 0 is unbounded if and only if Ol is infeasible. Dually, if Ol is feasible, then O'- is unbounded if and only if 0 is infeasible.


103

Proof. ad (i) : Let X E 0 be feasible. Then, by definition, X is optimal if and only if there is no Z E 0 which is improving with respect to X. Perhaps the situation is most easily understood if we illustrate it as follows: Suppose that the elements of E have been reordered

... = X = 0

such that Xl = ... = Xk = + and Xk+1 =

(0 0. For sufficiently large A > 0, either xa or x" is not in K+x. Hence C cannot have extreme rays. Now let C be pointed, i.e. ker B = 0. This means that the map-

ping: x - y = Bx is 1 - 1, implying that x E C is a proper convex combination of x', x" E C if and only if y = Bx is a proper convex combination of y' = Bx' and Y" = Bx". From this one concludes that x is extreme if and only if it is elementary. Thus in particular, C does have extreme vectors in this case. 0 After these preliminaries, let us turn to MINKOWSKI's Theorem:

116

CHAPTER 7. BASIC FACTS EV POLYHEDRAL THEORY

Theorem 7.6 (MINxowsKI) Every pointed polyhedral cone is the conic hull of its (finitely many) extreme rays.

Proof. Let C = P(B, 0) be pointed, and let E C C be such that {K+e I e E E} is the set of the extreme rays of C. We claim that C = cone E.

"J" is trivial. "C": Let x E C, and let y = Bx. Then y < 0 is the conformal sum of elementary vectors y' E im B, i.e. y = yl + yk, k _> 0. Let x' be corresponding elements of C, i.e. y' = Bx' (i = 1, , k). Then the xi's are elementary, and since C is pointed, they are extreme. Since C is pointed, the mapping x -+ y = Bx is 1 - 1, implying that x = E x. This shows that x E cone E. 0

Corollary 7.7 Every polyhedral cone C can be written as a conic hull of finitely many vectors. More precisely,

C = cone E+ linealC

for a finite setECIK". Proof. This follows from MINKOwsKI's Theorem and Lemma 7.4. Note that this actually is a conic hull representation since lineal C = cone(E' U -E') where E' is any basis of lineal C. 0

7.2

Polarity

Duality between vectorspaces is not a really difficult theory, and the Duality Theorems it provides are not very hard to understand, too (once you are used to work with them). From a geometrical point, however, our understanding is still very poor. If we try to imagine a dual pair of subspaces (L, Ll) in K" then most of us will have to content themselves with n = 3. (There exist people, who claim that they have no problems with imagining higherdimensional spaces. In case, they are mathematicians, they are usually called "geometers".) The loss of intuition, due to the above mentioned widespread inability of human race, may sometimes be compensated (to some extent) by associating low dimensional polyhedral cones to high dimensional subspaces. More precisely, to every subspace L = im B < K', there

7.2. POLARITY

117

are two cones associated in a natural way, namely C = P(B, 0) and CP = cone B. The former is polyhedral by definition and the latter will be proven to be polyhedral below. An obvious relation between these two is the following: If X E C and y E CP then xT y < 0.

Definition 7.8 Let S C IK". Then Sp :_ {y E 1K" I yTS < 0 bas E S}

is called the polar cone of S, or simply the polar of S. In other words, y E SP if and only if yTS < 0 is valid for S, i.e. SP can be regarded as a cone of "valid inequalities".

Lemma 7.9 (i) For S C ]K", SP = cone SP = (cone S)p and SPP D S. (ii) For S, T C IK", (S U T )p = SP n TP.

(iii) For a subspace L < IK", LP = L'. More generally, if S C 1K"

then Sl = SP n (-S)P. Proof. This is left to the reader as a straightforward exercise. 0

The following turns out to be essentially a restatement, or a polyhedral interpretation of FARKAS' Lemma:

Proposition 7.10 The polar of C = P(B, 0) is CP = cone BT and vice versa. Hence, in particular, C = CPP for every polyhedral cone.

Proof. Let E denote the set of rows of B. Then by Lemma 7.9 (i), cone(BT )P = cone(E)P = EP = P(B, 0), i.e. P(B, 0) is the polar of cone BT. Next, let us show that cone BT = P(B, 0)P.

"C" is trivial.

"D": Let x E K" and x V cone BT = cone E. By Theorem 4.9, x can be separated from E, i.e. there exists u E K" such that Bu _< 0 and xTu > 0. Hence xTu > 0 for some u E P(B, 0), i.e. x V P(B, 0)P. 0

Corollary 7.11 If C C IK" is a polyhedral cone, then lin CP = (lineal C)'.

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CHAPTER 7. BASIC FACTS IN POLYHEDRAL THEORY

Proof. Let C = P(B, 0). Then CP = cone BT, hence lin CP = lin BT = im BT and lineal C = kerB. 0

Note that, by Proposition 7.10, FARKAS' Lemma may be restated

as follows: For every y E K", yTx < 0 is valid for C = P(B, 0) if and only if y E CP = cone BT. This may help us to get a better understanding of FARKAS' Lemma sometimes: Given L = im B with B E Klox3, say, then the "linear" version of FARKAS' Lemma (Theorem 4.6) is a statement about L and Ll being subspaces of K10, while the above version relates C and C°, which are both contained in W. This "polyhedral" point of view of a subspace L < K" (or, more generally, an oriented matroid 0) will be emphasized throughout in what follows. An easy consequence of MINKOWSKI's Theorem is the following:

Corollary 7.12 If C C En is a polyhedral cone, so is its polar. Proof. By Corollary 7.6, C can be written as cone E for some finite set E C K". Let B denote the matrix whose rows are the vectors in E. Then C = cone BT and hence CP = P(B, 0) is a polyhedral cone. 0

A further immediate consequence is a homogenized version of WEYL's Theorem, which we state separately:

Theorem 7.13 (WEYL) For any finite set E C IK's, coneE is a polyhedral cone.

Proof. By Proposition 7.10, cone E is the polar of a polyhedral cone, and by Corollary 7.12, this is a polyhedral cone again. 0

Note that we obtained a proof of WEYL's Theorem by combining Proposition 7.10 and Corollary 7.12. These, in turn, are essentally due to FARKAS' Lemma and MINKOWSKI's Theorem, resp. On the other hand, WEYL's Theorem implies both FARKAS' Lemma (in fact it strengthens Corollary 4.12) and MINKOWSKI's Theorem. The latter may be seen as follows: As WEYL's Theorem implies FARKAS' Lemma,

it implies Proposition 7.10, hence, in particular, C = C" for every polyhedral cone C = P(B, O). Since CP = cone BT is a polyhedral cone by WEYL's Theorem, we may write CP = P(A, 0) for some matrix A, hence C = CPP = P(A, Of = cone AT, which is MINKOWSKI's Theorem. Summarizing, we might say that WEYL equals FARKAS plus MINKOWSKI. Of course, this statement does not mean anything, since

all three are valid in K. However, this can be made precise in a more

7.3. FACES OF POLYHEDRAL CONES

119

general setting, in which WEYL's Theorem is equivalent to FARKAS' Lemma plus MINKOwSKI's Theorem, but neither of them holds in advance. (Essentally this more general setting arises by replacing K by a more general structure, e.g. a ring or a semiring, cf. [38].)

7.3

Faces of Polyhedral Cones

In Section 7.1 we introduced the notion of extreme rays: If C C_ K" is a polyhedral cone and F = K+e is an extreme ray of C then F is a polyhedral cone which may be considered as an "extreme subcone" of C. Generalizing this notion, we may define extreme subcones of C in the following way:

Definition 7.14 Let C C 1h be a polyhedral cone. Then a polyhedral

cone F C C is called an extreme subcone or a face of C if the following holds: If x E F is a proper convex combination of x', x" E C

then x', x" E F. (Since F and C are cones, this amounts to say that

x = x' + x" E F implies x', x" E F.) A face F is called proper if F C C. A maximal proper face of C is also called a facet of C. The dimension of a face F is defined to be dim F := dim(lin F).

Example 7.15 (1) Let C = 112. Then F = 1K+ x 1K C C is a polyhedral cone, but not a face of C. In fact, the only face of C is C itself. (2) Let C =1K+ x K. Then the faces of C are {0} x K and C. They have dimensions equal to 1 and 2,resp.. (3) Let C = 1Ffa+. Then the faces of C are {0}, 1K+(o), 1K+(1), and C. Their dimensions are, resp., 0,1,1 and 2. The following two results are straightforward consequences of the definition of faces.

Lemma 7.16 Let C be a polyhedral cone, and let .T denote the set of faces of C. Define a relation " 0. Note that, by convention, if P = {x} consists of a single point, then int P = P = {x}. The boundary of a polyhedron P, denoted by 8P, will always mean the boundary of P considered as a subset of L = of P. Thus OP = P \ int P, since P is a closed subset of L. The definition has been given for general polyhedra P C K" rather

than just polyhedral cones only for further use in Section 7.8. For the time being, we will deal with polyhedral cones only. Note that if C C_ K" is a cone, then afC can be replaced by lin C. The above definition gives a topological characterization of interior points. There is an equivalent algebraic characterization which we are going to derive next.


122

Definition 7.23 Let C = P(B, 0) for some B E 1K"""". If x E C then

Q(x) := {i E {1, ... , m} I B;.x = 0}

is called the equality set of x. More generally, if S C C, then

Q(s) _ (1 Q(x) xES

is called the equality set of S.

Lemma 7.24 Let C = P(B, 0). Then x E C is an interior point of C if and only if Q(x) = Q(C).

Proof. "=": Let x E int C. Then Q(x) 2 Q(C) by definition, thus we are left to show that the reverse inclusion holds, i.e. that Q(x) C Q(x') for every x' E C. Thus let i E Q(x) and x' E C. Then u = x - x' E lin C. Since x E int C, this implies x + eu E C for some

e > 0. Thus, in particular, 0 > B;.(x + eu) = eB;.u = -e&.x'. Hence B;.x' > 0 and since x' E C this implies i E Q(x').

Suppose Q(x) = Q(C). Let u be an element of the unit ball in lin C, i.e. u = x' - x" for some x', x" E C and 1jull < 1. Then Q(x) = Q(C) C Q(x')f1Q(x"). Thus, whenever we have B;.x = 0 for some i then B;.u = 0. Choosing e > 0 small enough, we get x +eu E C for every u in the unit ball in lin C. Thus X E int C. 0

Corollary 7.25 Let C C 1K' be a polyhedral cone. Then int C :A 0.

Proof. Write C = P(B, 0). It follows from the definition of equality sets that for every i 0 Q(C) there exists some x' E C such that i ¢ Q(x'), i.e. B;.x' < 0. Then obviously t = Ei%Q(c) x' is an interior point of C. (Note that the empty sum corresponds to 2 = 0 by convention.) 0

Now let us turn to the description of faces in terms of the matrix B. As a first step we prove the following description of linear hulls:

Lemma 7.26 Let F be a face of C = P(B, 0). Then

linF={uE1K" I B;.u=0 ViEQ(F)}. Proof. "C" is immediate from the definition of Q(F).

7.4. FACES AND INTERIOR POINTS

123

"D": Let u E K" such that Bi.u = 0 V i E Q(F). Let x E int F. Then B(x+eu) < 0 fore > 0 sufficiently small. Hence x':= x+eu E C. Similarily, x" = x - Eu E C for e > 0 sufficiently small. Since x is a proper convex combination of x', x" E C, and F is a face, we conclude that x', x" E F. Thus u = A(x' - x") E lin F. Now we are ready to prove our main theorem, describing faces in terms of B.

Theorem 7.27 Let C = P(B, 0) be a polyhedral cone. Then F C C is a face of C if and only if

F=Cn{uElK"1Bi.u=O ViEQ(F)}. Proof.

is an immediate consequence of Proposition 7.21 and

Lemma 7.26.

"G" is straightforward from the definition of faces. Let F be as assumed in the claim. Then F C C is a polyhedral cone. Furthermore,

suppose that x = x' + x" E F for some x', x" E C, i.e. Bx' < 0 and Bx" _< 0. Then for every i E Q(F) we have 0 = Bi.x = Bi.x'+ BB.x" < 0, implying that x', x" E F. This shows that F is a face. The above theorem is sometimes stated in a slightly different form:

Definition 7.28 Let C C IK" be a polyhedral cone, and let cx < 0 be a valid inequality for C. Then H = cl is called a supporting hyperplane of C. (Note that c = 0 is not excluded.) Corollary 7.29 F C_ C is a face if and only if F = C n H for some supporting hyperplane H of C.

Proof. Let F be a face of C. Then

F=Cn{uEK"IBi.u=O ViEQ(F)}. Let c := EIEQ(F) Bi., and H := cl. Then H is a supporting hyperplane of C and F = C n H. Let H = cl be a supporting hyperplane of C such that F = C n H. Then C \ F = {x E C I cx < 0} is a convex set. Furthermore, lin F= lin C n H, hence C n lin F= C n H = F. Thus F is a face by Proposition 7.21.

124

7.5

CHAPTER 7. BASIC FACTS IV POLYHEDRAL THEORY

The Canonical Map

The most important aspect of Theorem 7.27 is that it allows us to study the faces of C = P(B, 0) in terms of the linear space im B or, more precisely, in terms of 0 = o'(im B). In fact, consider the map

aoB:K"-+ 0. This will be called the canonical map from K" to 0. Proposition 7.30 The canonical map induces a 1-1 correspondence between the faces of C and the set {Y E 0 Y < 0}. If F is a face of C, the corresponding sign vector is given by Y = a o B(int F). If Y E 0, Y < 0, the corresponding face F of C is given by F = {x E IK" I a o B(x) -< Y}, i.e. F = (a o B)-1[Y1. This correspondence is

such that F < F' (cf. Lemma 7.16) if and only if Y -< Y' for the corresponding sign vectors.

Proof. By Theorem 7.27, F is a face of C if and only if

F=CfI{xE]K"IBB.x=O ViEQ(F)}. Thus there is a 1-1 correspondence between faces F of C and their equality sets Q(F). On the other hand, every equality set Q(F) corresponds to a unique Y E 0 such that Y < 0 and Y° = Q(F).

Thus the -< relation on 0 may be interpreted as "being a face of ...". Of course, so far, we have only considered a small part of 0, namely the set {Y E 0 1 Y _< 0}, i.e. the set {Y E 0 1 Y -< (...... -)}. But the results are easily extended to the whole set 0: Definition 7.31 A cone C C Iff" is called definable from Bx < 0 if and only if

C={xE1K"I aoB(x)-< X} for some X E 2±E. Alternatively, a cone is definable from Bx < 0 if its defining system of inequalities is obtained from Bx < 0 by replacing some of the

inequalities "" or "_" . Theorem 7.32 The canonical map induces a 1-1 correspondence between cones definable from Bx < 0 and the sign vectors in 0. If F and F' are two cones definable from Bx < 0, and Y, Y' are the corresponding sign vectors then F < F' if and only if Y -< Y'.

7.5. THE CANONICAL MAP

125

Proof. This is a straightforward extension of Proposition 7.30, obtained by replacing some of the inequalities B;.x < 0 by B;.x > 0 or B;.x = 0. Theorem 7.32 provides, in a sense, a geometrical (or topological) realization of a linear OM 0 = a(im B). The purely "combinatorial" structure of the poset (0, - .F(CP) is given by

F-+CPnF1. Proof. Let cp be as defined in the claim. We will show that cp is an antiisomorhism between .F(C) and .F(CP) in several steps.

(1) cp :.F(C) -> .F(C"): Let F be a face of C. Then every c E F defines a valid inequality cTx < 0 for CP. Hence CP n cl is a face of CP. Therefore,

w(F) = CP n F-L=n{CPncl I c E F} is an intersection of faces of CP, and hence is itself a face of CP.

CHAPTER 7. BASIC FACTS IlV POLYHEDRAL THEORY

130

(2) F < F' implies w(F) > cp(F'): This is obvious from the definition of W.

(3) cp is injective: Define cp :.F(CP) --+.F(C) by

F-+CnF1. Note that it follows from 1) that cp actually maps .F(C") to .F(C). Now 3) will follow, once we have shown that cp o cp = id.

To see this, let F E .F(C). Then, by Lemma 7.9 and Corollary 7.11, we get

(CP n F1)1

(-CP n F-L)P n (CP n F-L)P ((CPU -CP) n F-L)P (cone(CP U -CP) n F1)"

(lin(CP) n F-L)P = ((lineal C)1 n F-i)"

(F1)" = F11 = lin F Hence cpcp(F)

= C n (CP n F1)1 = C n lin F = F.

(4) cp is surjective: We have just seen that cpocp = id. By symmetry (i.e. replacing C by C"), we conclude that cpocp = id, too. Hence cp is surjective.

This completes the proof of the theorem. 0

Lemma 7.42 Let C C IK" be a polyhedral cone, and let CP its polar. Then the antiisomorphism

cp:=F-iF=CPnF1 is such that dim F + dim .P = n for every face F of C.

Proof. It suffices to show that lin F = F1 for every face F of C. To see this, first note that, as in step (3) in the proof of Theorem 7.41,

(CP n F1)1 = lin F. Hence

lin F = lin(CP n F1) = (CP n F1)11 = (lin F)1 = F1 . 0

7.8. GENERAL POLYHEDRA

131

Corollary 7.43 If C C IK" is a polyhedral cone, and F and F are two faces of C such that F is a facet of F, i.e. F covers F in the face lattice of C, then dim F = dim F- 1. Proof. W.l.o.g., suppose that F = C, i.e. that F is a facet of C. Since w:= F - P is an antiisomorphism, this implies that 0, the minimal face of CP, is a facet of P. Thus, by Lemma 7.42, we are left to show the following: If C C K" is a polyhedral cone with minimal face F and a face F' covering F, then dim F = dim F- 1. This is obvious in case C is pointed, for in this case F = {0}, and Fmust be an extreme ray K+e of C. The general case, however, is easily derived from the special

case of pointed cones. In fact, let C C K" be an arbitrary polyhedral cone and consider the associated pointed cone Co = C fl lineal C1. Then F is a face of C if and only if F = F0 + lineal C for some face

F0 of CO, implying that dim F- dim F = dim F,'- dim F0 for any two faces F, F of C. This proves the claim. 0

Theorem 7.41 states that the face lattices .F(C) and .F(C") of a polar pair of polyhedral cones are, in a sense, the same thing. Now consider the oriented matroid 0 = a(im B) and the set T = {Y E 0 Y < 0}. We know already that T and .F(C) are isomorphic. Hence T and.F(CP) are antiisomorphic. Hence both.F(C) and.F(CP) can be considered as "geometrical realizations" of the abstract, purely "combinatorial" lattice (T, ). When we study such "tope lattices" in a more abstract setting in Chapter 8, it will be helpful to associate to T intuitively either one of the two possible "realizations" C and CP (although, in general, i.e. when 0 does not arise from a subspace of K", such realizations may not exist). It does not matter at all, whether we think of C or CP when working with T as above, however, it will be less confusing to fix one of them arbitrarily. We decided to fix C. Thus given 0 = a(im B) and T as above, we will always interprete the order -< on T as the inclusion order on the faces of C = P(B, 0).

7.8

General Polyhedra

It is not difficult to generalize the results of the preceding sections for polyhedral cones to general polyhedra. It is not very interesting either, but we would like to do so, just for the sake of completeness. Of course the basic tool to be applied is (de-) homogenization. Let us start with defining extreme points and faces of general polyhedra.

132


Definition 7.44 Let P C IK" be a polyhedron, and let C C 1K"+1 denote its homogenization. Then x E P is called extreme if the corresponding vector (7) E C is extreme. Extreme points of polyhedra are

also called vertices.

It is straightforward from the definition that x E P is a vertex if and only if x is not a proper convex combination of x', x" E P.

Example 7.45 1. Let P = lineal P be a subspace of W. Then P has no vertices.

2. LetPC1Kbedefined byP:={xE1K10<x n of vertices (cf. the references in Section 7.9).

Definition 7.53 Let P C En be a polyhedron, and let F = F(P) denote the set of faces of P. For F, F' E .F, let F < F' if F is a face of F' (or, equivalently, F C F'). Then (.F, 0, i.e. T = (...... +). (In fact, reorientation induces an isomorphism of 0 as a poset.) Let f, f' denote the facet equators of [Y] and [Y'], resp. The proof is by induction on IEl. If E = { f, f'}, the claim is obviously true, since

CHAPTER 8. THE POSET (0, - 1. Then there exists a tope [T'] E L such that L':= L \ {T'} is again a lump. Furthermore, we have [C] n [Ti = O[c'] n O[T J = [u(T', T)]

(Here, of course, [G'], the "closure" of L, is to be understood in the same way as the "closure" of an umbrella.) Proof. The first claim is obvious. In fact, removing any tope T' E G with sep(T, T') = r leaves a lump C' := G \ [T'] of radius r or r - 1. Now let us show that if T' and G' are as above, then [Cl n [T'] = a[C'] n 8[T'

8.4. CONSTRUCTIBILITY OF 0

165

"3" is trivial. To prove the converse inclusion, let Z E [L'] n [T']. Hence Z - T' and Z -< T" for some tope T" E L. Since sep(T, T') and sep(T, T") are both less or equal to r, we obviously have sep(Z, T) =:

i < r. Thus S := Z o T E V. Choose any path from T to S and append any path from S to T', both paths being constructed as in Proposition 8.16. This results in a path T =To,T1,...,T,, = S, T,+1,..., T, = T'.

Note that Isep(T,TT)l = j for all j, hence T1i...,Tr_1 E V. Along the path from T to S we "traverse" all elements e E sep(S, T). (We assume that 0 is simple, hence we traverse exactly one element e E E at a time.) Along the path from S to T' we only traverse all elements

in sep(T,T') \ sep(T, S). From this it is clear that Z conforms to all topes Ti, T11,.. . , T,. In particular Z -< Tr_1. Thus, if [Y] denotes the common facet of Tr_1 and T,. = T', then Z E [Y]. But [Y] obviously is an element of a[L'] and i9[T] (recall that Tr_1 E L'). Hence Z E ,9[L'] n 3[T'], which proves the inverse inclusion.

Note that our above argument actually shows that if [Z] C [L'] n [T'], then [Z] C_ [Y], where [YJ is a facet of [T'] with facet equator [U(T',T)]. e E sep(T',T), i.e. [Y] E U(T',T). Hence [L'] n [T'] The converse inclusion is clear, since every facet [Y] C [U(T', T)]

has a facet equator e E sep(T', T). Hence gT' E L', showing that [YJ E [L'] n [Ti. Thus we get

[L'J n [TI = [U(T', T)] a

Theorem 8.38 Let 0 be an OM-poset of dimension n. Then (i) If L is a lump of radius r < IEJ, then [L] is B-shellable of dimension n.

(ii) 0 is S-shellable of dimension n. (iii) 6 = 0 U {1o} is B-shellable of dimension n + 1.

Proof. (i): Let L be a lump centered at T of radius r < JEl (i.e. -T 0,C). If ILI = 1, then [L] = [T] is B-shellable of dimension n by Theorem 8.35. If ELI > 2, then, by Lemma 8.37, we find T' E L such that L' = L \ [T'] is a lump and [LI n [TI = a[v] n a[T'] = [U(T', T)]

CHAPTER 8. THE POSET (O, - 0 induces a 1-1 correspondence between the cells

[Y] C 0 and the corresponding cones Cy C R" such that X Y in O if Cx < Cy, i.e. Cx is a face of Cy. Thus in case 0 is a linear OM, i.e. 0 = o(im B) for some matrix B, the purely combinatorial structure 0 can be "realized" by a topological structure consisting of a family (Cy I Y E 0) of polyhedral cones as above. The purpose of this chapter is to present a similar kind of topological realization for general OMs. Section 9.1 revisites the canonical map o o B and slightly modifies the above "realization", introducing the notion of "linear sphere systems" in order to prepare the more general concepts to be defined in the sequel. Section 9.2 presents an example of a nonlinear OM, i.e. one which cannot be written as 0 = o *m B) for any matrix B. Section 9.3 introduces the concept of (general) sphere systems as a topological analogue of the linear sphere systems of Section 9.1.

9.1

Linear Sphere Systems

Let B E RE"" and let 0 = Q(im B) the correspoding OM on E. Recall from Section 7.5 that the canonical map o o B : R" - 0 gives a 1-1 correspondence between cells [Y] C 0 and polyhedral cones in R", definable from Bx < 0. For Y E 0 let Cy C R" denote the cell (i.e. the polyhedral cone) corresponding to [Y], i.e. Cy = (o oB)-'[Y]. We have indicated already in Section 7.5 that the "combinatorial

CHAPTER 9. TOPOLOGICAL REALIZATIONS

172

structure" of 0 is reflected in the geometrical (or topological) struc-

ture of the system S = (Cy I Y E 0) in some sense. In particular, the "combinatorial" relation -< in 0 corresponds to the "topological"

relation < (i.e. "is a face of") in S. In this section we will analyze the topological structure of the system S (and a related system to be defined later on). The results are not very surprising and we state them only in order to prepare the definition of a more general class of "cell systems" in Section 9.4.

Proposition 9.1 Let S = (CY I Y E 0), and for each Y E 0 let s(Y) := int Cy. Then the following holds: (i) s(O) is a subspace of 1R".

(ii) 1R" = O{s(Y) I Y E 0}. (iii) For every Y # 0 the set Cy = U{s(X) I X E [Y] } is a polyhedral cone with aCy = U{s(X) I X E a[Y1}. Proof. (i) is trivial, and (ii) and (iii) are proved in Theorem 7.32. 11

One can show that every polyhedral cone C C R" -provided C lin(C)- is homeomorphic to a closed halfspace H+ C Rd where d = dim C. Thus, from a topological point of view, Proposition 9.1 provides a relationship between 0 and a system of "topological halfspaces" satisfying conditions (i)-(iii) above. In the following we are going to replace the system (Cy I Y :A 0) of topological halfspaces by an "equivalent" system of topological balls. This is not a very essen-

tial step, in fact, it is more or less only due to tradition and the fact that balls are, maybe, somewhat handier than halfspaces.

Definition 9.2 A topological n-ball, or, for short, a topological

ball is a topological space B, homeomorphic to the n-dimensional unit ball B = {x E 1R" IIxII < 1}. The number n is called the I

dimension of B. Similarily, a topological n-sphere, or, for short, a topological sphere is a topological space S, homeomorphic to the n-dimensional sphere S" = {x E R"+1 I IIxII = 1}. The number n is called the dimension of S. By convention, IR° = {0}, thus S-1 = 0 and B° = {0}. If B is a topological ball and h : B" -> B is any homeomorphism then the boundary of B is defined to be aB := h(aB") = h(Sn-1)

and its interior is defined to be int B := B \ B. The boundary of a topological sphere S is defined to be aS := 0 and its interior is

int S:= S.

9.1. LINEAR SPHERE SYSTEMS

173

Remark 9.3 Note that the boundary 8B of a topological ball B is in fact well-defused: If h : B -+ B and g : B," - B are homeomorphisms then g-1h : B" -> B,,, is a homeomorphism. From this one concludes that m = n and g(BB,,,) = h(BB,,,) by means of the Domain Invariance Theorem (cf. Section 1.3).

Lemma 9.4 Let C C 1R" be a polyhedral cone. Then C n Sn-1 is a topological sphere or a topological ball, according to whether or not C is a subspace of 11Z".

Proof. Elementary proofs of this result are straightforward but somewhat tedious. We leave it as an exercise. Shorter proofs can be obtained by applying the PL-techniques which we are going to sketch in Section 9.4.

Now we are ready to replace our original system (Cy I Y E 0) by the system (CynSn-1 I Y E 0). Instead of considering the partition of Rn induced by the system of hyperplanes H° = {x E R" I Be.x = 01, we consider the partition of Sn-1, induced by the system of "hyperspheres" H° n Sn-1, e E E.

Definition 9.5 Let L < 1R" be a subspace. Then S := L n Sn-1 is called a linear sphere. If S = L n Sn-1 is a linear sphere and c E 1R" \ {0} such that H° := {x E S I cx =0} # S then H° is called a

linear hypersphere of S. H+ := {x E S I cx > O} and H- := {x E S I cx < 0} are called the sides of H° in S. The triple (H°, H+, H-) is

then called an oriented linear hypersphere of S. A linear sphere system is a family (H°, H.+, He )eEE of oriented linear hyperspheres

of Sn-1. The sets H: := Ha U H° and Ha := Ha U H° are called closed halfspheres. If A C E then F= n{H° I e E Al is called a flat of the system. The intersection of any set of closed halfspheres which is not a flat is called a supercell. A supercell which is not a proper union of two or more other supercells is called a cell. From Proposition 9.1 and Lemma 9.4 we get immediately:

Theorem 9.6 Let (H°,

H.-).EE be a linear sphere system. Then

the following holds:

(i) Any flat is a linear sphere. (ii) If F is a flat and F ¢ H° then H° n F is a linear hypersphere of F with sides HH n F and He n F. (iii) Any supercell is a topological ball.

174


Proof. (i) and (ii) are trivial, and (iii) is the content of Lemma 9.4. 0 Of course, almost nothing has happend so far. Essentially, all we

did was to replace the hyperplanes {x E R" I B..x = 0} by the linear hyperspheres {x E Sn-1 Bt.x = 0} and the cones Cy by their corresponding cells Cy fl S"-1. But, as we noted already, the I

main purpose of our work was to prepare the introduction of (general) sphere systems in Section 9.3.

9.2 A Nonlinear OM A natural question that comes up as soon as oriented matroids are defined, is whether we really got something more general than just systems of sign vectors arising from linear subspaces. Now the time has come to answer this question: There do exist oriented matroids that can not be written as 0 = a(im B) with B ERE". We do not intend to give a rigorous proof of this (which would be rather lengthy), but just sketch an example of one such oriented matroid. Consider a system of ten lines in the plane R2 as indicated below:

This is called "DESARGUES Configuration" and a wellknown theorem from geometry ("DESARGUES Theorem") says that in any such

configuration the three lines aa', bb' and cc' meet in a point p. Now

9.2. A NONLINEAR OM

175

imagine this system of lines (with the three lines meeting in p) being embedded into the 2-dimensional sphere S2 C R3 (with the lines replaced by 1-dimensional linear spheres), thus giving rise to a linear sphere system as introduced in Section 9.1. The corresponding oriented matroid is a linear one, which can be written as 0 = a(im B) for some B E R10"3 Recall that every Y E 0 corresponds to a cell in the linear sphere system (i.e. Y represents a "possible" position of a point on S2 with respect to the system of hyperspheres). Now we slightly deform one of the three lines meeting at p, thereby making it pass by the point p. This gives rise to a new system of "lines" on S2 which looks like indicated below:

Figure 9.2

cl

Of course, the corresponding operation has to be performed at the antipodal point -p as well. This gives rise to a new system of "cells", i.e. possible positions of a point on S2 with respect to the modified system of lines. Thus, it is intuitively clear, how to derive a system of sign vectors from this, each sign vector representing a possible position of a point on S2 with respect to the modified system of lines. One can show by brute force that the system of sign vectors arising that way is an oriented matroid O. We won't do so, however, since we are going to prove a more general result in Section 9.3, which says that any such system of topological lines (or more generally "hyperspheres") on a sphere gives rise to an oriented matroid. The above oriented matroid 0 is called the "Non-DESARGUES Matroid". This cannot be written as 0 = a(im B) for any matrix B.


176

In fact, if 0 were linear, this would imply that 0 can be represented by a linear sphere system, contradicting DESARGUES Theorem.

9.3

Sphere Systems

As we have indicated in Section 9.2, an OM can in general not be "realized" by a linear sphere system. In this section we will introduce a more general class of sphere systems which will turn out to be large enough to allow realizations of general (nonlinear) OMs. This generalization essentially consists in replacing the linear spheres by topological ones.

Definition 9.7 Let S be a topological n-sphere. If h : Sn -- S is any homeomorphism, then H° = h(Sn-1 x {0}) is called a hypersphere of S. The triple (H°, H+, H-), where H+ and H- are the two components of S\ H°, is called an oriented hypersphere of S. H+ and H- are called the sides of H° in S. We write H+ := H+ U H° and H- := H- U H°. These are the closures of H+ and H-, resp. in S. Therefore, H+ and H- are also called the two closed halfspheres associated to H°. Now consider any triple S = (S, E, 7i) where S is a topological sphere, E is a finite set, and l is a family ((H°, H.+, He) I e E E) of oriented hyperspheres of S. If A C E then F := n{H° I e E Al is called a flat. The intersection of any set of closed halfspheres which is not a flat is called a supercell. A supercell which is not a proper union of two or more supercells is called a cell. If K is a supercell, then F= n{ H° I K C_ H°} is called the flat generated by K. This is the smallest flat containing K. S is called a sphere system, provided the following holds: (i) Any flat is a topological sphere.

(ii) If F is a flat and F g H° then Fn H° is a hypersphere of F with sides F fl H. and F fl H.-. (iii) Any supercell is a topological ball.

Theorem 9.6 states that every linear sphere system is a sphere system. Furthermore, it is intuitively clear that the lines we used to describe the nonlinear OM of Section 9.2 give rise to a sphere system. We will not analyze this here in detail, because we are going to prove in Section 9.4, that, in general, every OM can be "realized" by a sphere system. In this section we will only show that, conversely, every sphere system "realizes" an OM. (A precise definition is straightforward, but

9.3. SPHERE SYSTEMS

177

will be given below.) Before we can proceed, however, we will have to investigate the topological structure of sphere systems.

Lemma 9.8 Let S

(S, E, H) be a sphere system and let f E E.

Define

fl/f Slf

eEE\f} and

(S,E\f,f/f)

Then S/f is a sphere system again. Furthermore, one can define a system S \ f in the obvious way as follows: Let §:= Hf (which is a topological sphere) and let

E:= {eEEI HO $,§j By axiom (ii) for sphere systems every

H°:=H°f

eEE

is a hypersphere of S with sides

Ha :=He flS

and

fl.- .=H; f1

Let

fl \ f :=

I e E E) and

S\f :=

Then S \ f is a sphere system again. Proof. This is left to the reader as an easy exercise. 0

Remark Be sure that you have well understood the meaning of Definition 9.7. In particular, (iii) states that every supercell K, considered as a topological subspace of S, is homeomorphic to some ddimensional ball Bd C W. The topology of K is that induced by S, which is the same as that induced by F, where F is the flat generated by K. (This is due to transitivity of induced topologies, cf. Section 1.3).

Since every supercell K is a topological ball, its interior int K is defined according to Definition 9.2. For the moment, however, let us consider intFK, the interior of K within the flat F it generates. (This will turn out to be equal to int K in a minute.)


178

Lemma 9.9 Let K be any supercell, say,

K= n{H: I e E Al for some subset A C E and let F denote the flat generated by K. Then

intFK=n{H: IeEA, K¢H°}nF.

Proof. Let e E A such that K g H.O. Hence, in particular, F¢ H.O. Thus H: n F is open in F. This shows that

U:=n{H:IeEA, K¢H°}nF is open in F and contained in K. Hence U C intFK (which, by definition, is the largest F-open subset of K). On the other hand, if V is open in F and V C K, then, in particular V C_ F n H for every e E A. Since V is open in F, we further conclude that

VCintF(FnHf)=FnH for every eEA. Hence

VCn{H I eEA}nFCU.

In particular, this holds for V := intFK. Hence U = intFK, q.e.d.. 0

Proposition 9.10 Let K be a supercell, say, K= n{Ht , e E Al for some A C E, and let F denote the flat it generates. Then the following hold: (i)

intFK 00

(ii) intFK is connected, i.e. it cannot be partitioned into two disjoint F-open sets.

(iii) c1F(intFK) = K

(iv) dimK = dimF (v) intFK = int K

(vi) F=n{H°1eEA, KCH°} Proof. We will first prove (i)-(v). Write

K=n{Ha IeEA}nF. W.l.o.g. we may assume that K H° for every e E A. The proof of (i)-(v) will be by induction on k := IAA. For k = 1, i.e. A = {e},

9.3. SPHERE SYSTEMS

179

the claims (i)-(v) follow immediately from the fact that H° fl F is a hypersphere with sides H H fl F and H. fl F. Assume now that k > 2. Let a E A and consider

M:=n{HV IeEA\a}f1F.

Thus KCMCF.IfMwereaflat,then M=FandK=Ha flF, contradicting our assumption that IAA > 2. Hence M is a supercell

and the flat generated by M is F. Furthermore, since M D K, we have that M fg H° for every e E A \ a. Hence, by induction, (i)-(v) hold for M. From this we will conclude that (i)-(v) also holds for K.

ad (i): By Lemma 9.9, intFK= intFM fl H:. Assume that intFK = 0, i.e. intFM C Ha-. Since the induction applies to

M, we get M = cIF(intFM) C cIFHH = Ha-. But then K C M fl Ha Ha, contradicting our assumption. Thus (i) holds for K.

ad (ii)-(v): Choose X E F \ K. Since F is a topological sphere, there exists a homeomorphism h : F -+ S- C R7+1. Then F \ x -+ Sm \ h(x) R"` (by stereographic projection) gives a homeomorphism g : F \ x - R"` mapping K onto some subset B := g(K) C R!'. Now B, being a homeomorphic image of K, is of course a topological ball. Furthermore, we have intRm(B) = g(intFK) 96 0, hence B is in fact m-dimensional. Thus int K = g'1(intRmB) = intFK is connected, and its closure in F equals

g-1(B) = K. ad (vi): By definition, the flat generated by K equals

F=n{H°1eEE, KCH°}. We have to show that this equals

F:=n{H°1 eEA, KCH°}. Obviously, F C F, hence it suffices to show that dim F = dim F. This will be proved by induction on k :_ I E\AJ. If k = 0, there is nothing to show. Thus assume k > 1 and let f E E \ A. Consider the sphere system S/ f obtained by removing the hypersphere

HI as described in Lemma 9.8. K is (still) a supercell with respect to S/ f and, by induction, the flat it generates equals F. Hence (vi), applied to the sphere system S/ f, yields dim P = dim K. On the other hand, (iv) applied to our original system

S gives dimF = dimK, q.e.d..


180

0

Now we are ready to prove that sphere systems give rise to oriented matroids.

Definition 9.11 Let S = (S, E,1-l) be a sphere system. Define v by

(Q(x))e:_

1+ ifxEH: - ifxEH, to if xEH°.

Let O(S) := Q(S) U {0}.

Theorem 9.12 If S = (S, E,7-l) is a sphere system then O(S) is an oriented matroid.

Proof. Let 0 = O(S). Before proving that 0 satisfies the OM-axioms, let us make the following simple

Observation: Let K be an intersection of closed halfspaces, say,

K=n{Hi I eEA}. If Kn H: 96 0 for some e E A, then K is a supercell (and not a flat). Proof. Suppose Kn H.+:$ 0 but K is a flat. By axiom (ii) for sphere systems, K n H° is a hypersphere of K with sides K n H. and K n H,-,. Thus, in particular K n H. is nonempty. But this is impossible, since

KCH:.

Now let us verify the OM-axioms for O.

(i) X, Y E 0 implies X o Y E O: Let X, Y E O with supp(X) 2 supp(Y) and let x, y E S such that X = a(x) and Y = o (y). Let

K:=n{H, ix,yE f } nn {H. l x,yEH.}. Then K is easily seen to be a supercell. In fact, take any e E suppX, say e E X+. Then x E KnHH , i.e. KnHH is nonempty and hence our above observation implies that K is a supercell. Now let

M:=n{H! ix EH., yEH.} nn f vix EH., and consider N := M n K. This is a supercell again. Indeed, if M = K, there is nothing to show any more. Otherwise, if N C K, there exists some e such that, say x E HH and y E H,-. But then x E N n HH $ 0 and our observation again implies that N is a supercell. By Proposition 9.10, int N $ 0. Obviously,

any zEintNgives Z:=v(z)=XoY.

9.3. SPHERE SYSTEMS

181

(ii) 0 has the approximation property: Let X, Y E 0 and let x, y E S and K as above. Let e E sep(X, Y), say e E X+ n Y-. Let K1 := K n 7,+- and K2 := K n H.. Now x E Kl n H. and

y E K2 n H.. In particular Kl n H` and K2 n H` are both nonempty, implying that Kl and K2 are supercells. Hence both have nonempty interior. From Lemma 9.9 we get

int Kl = int K n HH and int K2 = int K n He . But int K is connected (cf. Proposition 9.10) and therefore can-

not be the union of two nonempty open subsets int Kl and int K2. Hence int K n H° must be nonempty. Now it is easy to see that any z E int Kn H° gives Z = o (z) as an approximation of X and Y on e. In fact, suppose that f E supp X U supp Y

does not separate X and Y, say f E X+ \ Y-, i.e. x E H) and y E H#. Thus Kn H) is nonempty since it contains x. In particular, K ¢ HIO and therefore Lemma 9.9 gives that int K C H)-. This means that z E Hj , i.e. Z1 = +.

(iii) X E 0 implies -X E 0: Let 0 $ X E 0 and choose x E S with a(x) = X. W.l.o.g. assume that X > 0. Consider

K := n{H. IxEHH} n n{H°1xEH°} and -K := n{H. IxEHH} n n{S°1xEH°}. From our observation, it is clear that K is a supercell.

182


Claim Let A C E, let F be a flat and consider

K

-K

n{H: I e E A} n F

and

n{Hi e E A} n F.

Then, if K is a supercell, so is -K. Moreover, if K generates

F, then -K generates F. Once we have proved the claim, it is immediate that any y E int(-K) gives Y = a (y) = -X as required. Hence we are left to prove the claim. Suppose K is a supercell generating F. If

d := dimF = 0, then F = {x, y}, K = {x} and -K = {y}. From this the claim is obvious. We proceed by induction on

d = dimF. Thus assume that d > 1. If k := IAA = 1, say A = {a}, the claim is trivially true. We proceed by induction on k. Thus let k _> 2. If K C Ha for some a E A, then F C_ Ha, since K generates F. Hence

K = n{H. I e E A\ a} n F

and

-K = n{H IeEA\a}nF, so the result follows by induction on k in this case. Hence assume

that Kn Ha 96 0 for every a E A. We claim that also 8Kn Ha $ 0 for every a E A. Indeed, assume that OK n Ha = 0 for some

a E A, i.e. 8K C Ha n F. Pick any b E A \ a. Then K C Hb , hence 8K C Hb n Ha n F, i.e. OK is contained in one half of Ha n F (we tacitely assumed that the H. n F, e E A, are pairwise distinct). But this is impossible since both 8K and Ha n F are spheres of dimension d - 1. Hence we get that in fact 8K n Ha 96 0 for every a E A. Now choose any a E A and

z E 8Kn H:. Since z E OK, we get z E Hb for some b E A. Now consider the sphere system S \ b and let F denote the flat generated by k:= K n Hb. Consider

k :=

-k

n{Hf e E A\ b} n .P e E A\ b} n F.

Since z E k n H:, our observation shows that k is a supercell.

Furthermore, since F C F n Hb, induction on d yields that -K is a supercell generating F. In particular, -K g Ha (since K ¢ Ha). This further implies -K ¢ Ha, i.e. -K n Ha t- 0. Summarizing, we have proved that K n Ha # 0 implies (-K) n Ha # 0. This shows that -K is a supercell generating F. 0

9.4. PL BALL COMPLEXES

183

Note the nice relation between the combinatorial properties of 0 and the topological properties of sphere systems. In fact, 0 is closed under taking sums because int K $ 0 for any supercell K. The approximation property corresponds to connectivity of int K, where K is a supercell and finally, O = -O is due to the existence of "antipodal points" in spheres.

9.4 PL Ball Complexes The main result of Section 9.3 states that any given sphere system S gives rise to an oriented matroid 0 = O(S) in a natural way. Our final goal in this section is to show that, conversely, given any oriented

matroid 0, we may construct a sphere system S such that 0 = O(S).

Definition 9.13 Consider a triple 1C = (P, X, s), where P is a poset with minimal element 0, X C 1Rfor some n, and s is a map which associates to every p E P a subset s(p) C X, called the space of p. If Q C P, let s(Q) := U{s(q) I q E Q}. IC is called a topological ball complex (or ball complex, for short) if the following two conditions hold:

(i) X=U{s(p)IpEP}. (ii) For every p E P \ {0}, s[p] is a topological ball whose interior is s(p), i.e. its boundary is s(8[p]). Hence, in particular, s(O) is a topological sphere.

For P E P the set s[p] is called a cell of IC. X is called the space of K, also denoted by s(1C).

It is easy to see that the intersection of any two cells s[p] and s[q] (with p :A q) can be written as a union of cells in the boundary of s[p]

and s[q]. In fact, let x E s[p] fl s[q] and let p' < p be minimal with x E s[p']. Then obviously x E s(p'). Similarly, let q' < q be minimal with x E s[q']. Then x E s(q'). Hence from condition (i), we conclude that p' = q' and the claim follows. Example 9.14 Let P = 10,p} consist of just two elements with 0
s'(Ki) be a PLH compatible with f. Let P1 C_ P be the poset of Kl . Let p E P \ P1 be minimal. Then 8[p] is all contained in P1. Now h may be extended to the subcomplex corresponding to P1 U p by means of Lemma 9.30. Continuing this way, we end up with an extension s(K) -* s(K'). (Note that the first step in the induction is provided by our assumption that s(0) is PLH to s'(0').)

Theorem 9.36 (Sphere Construction Theorem) Let P be

a

poset, and let P1 and P2 be two order ideals such that P = P1 U P2 and P1 n P2 = 8P1 =8P2. Furthermore, suppose we are given two PL-ball complexes Kl and Ka with posets P1 and P2, resp., whose spaces are PL n-balls. Then there exists a PL-ball complex K whose poset is P and whose space is the boundary of a (n + 1) -simplex (-hence a PL n-sphere).

Proof. Let e1, ... en+1 E R"+1 denote the (n + 1)-dimensional unit vectors. Let An = cony O, el, ... , en

,

0 = cony 0, el, ... , en, en+1 and 0 = conv{0, e1, ... , en, -en+1} Furthermore, let B1 := 9A+ \ into" and B2 := 80- \ inton. Obviously, B1 and B2 are PL n-balls, and B1 n B2 = 80n is a PL (n - 1)sphere.

Let Ki and K' denote the subcomplexes of Kl and K2, resp., corresponding to P1nP2 = 8P1 = OP2. Since sl(Kj) and s2(K2) are PL n-balls, their boundaries sl(Ki) and s2(Ka) are PL (n - 1)-spheres (cf.

Lemma 9.31). In particular, there exists a PLH h': sl(K'l) - s2(q).


192

In fact, since the identity map id : 9P1 -+ 8P2 is an isomorphism, Lemma 9.35 shows that h' may be choosen to be "cell-preserving", i.e. such that h'(sl(p)) = 32(p) for every p E Pl n P2. Since sl()C') and s2(1Cz) are PL (n -1)-spheres there exist PLHs

hl : Bl n B2 - s(Ki) and h2 := h'h1 : B1 n B2 -* s(Ka). These may be extended to PLHs hi : B; -+ s;(K;). Now we define a PL-ball complex K whose poset is P and whose space is Bl U B2 in an obvious way. If p E P;, we define the space of p to be h; 1s;(p). (If p E Pl n P2, the two definitions agree.) This gives obviously rise to a PL-ball complex with poset P and space Bl U B2, which is the boundary of a (n +1)-simplex. 0

Note that, so far, we have made no use of NEWMANs Theorem. NEWMAN's Theorem will, however, become crucial when we construct PL-balls by pasting balls along part of their boundary, as we are going to do now.

Theorem 9.37 (Ball Construction Theorem) Let P be a poset and let Pl and P2 be order ideals such that P = Pl U P2 and Pl n P2 = 8P1 n 8P2. Furthermore, suppose we are given two PL-ball complexes Kl and K2 with posets Pl and P2, resp., such that the spaces of Kl and K2 are PL n-balls and the spaces of their subcomplexes corresponding to Pl n P2 are PL (n - 1)-balls. Then there exists a PL-ball complex K whose poset is P and whose space is an n-simplex (hence a PL n-ball). Proof. Let 1C and )CI be the subcomplexes of Kl and K2 corresponding to P1nP2. Since s;(K;) is a PL n-ball, s;(8K;) is a PL (n-1)-sphere for i = 1, 2 (cf. Lemma 9.31). By assumption, sti(K;) C sti(8K;) is a PL (n -1)-ball. Let K;' denote the subcomplex of 8K; which corresponds to [8P; \ (P1 n P2)], i.e. the order ideal generated by 8P; \ (P1 n P2).

Then s(K;') is the closure of sj(8K;) \ si(K;), which is a PL (n - 1)ball by NEWMAN's Theorem. Furthermore, we must have 81C; = 8Ks', since the space of 01C; = K; U K;' is a Pl-(n - 1) sphere and hence has empty boundary.

Now, since sl(1() and s2(Ka) are PL (n - 1)-balls, there exists a PLH s(K4) -> s(1(). By the Isomorphism Lemma, this may be chosen to be cell preserving. Now let O n-1 .= cony 0, e 1 , . .. , en_1}1 O .= cony 0, el, ... , en-1, enl and

A-

conv{0, el, .... en-1, -en}.

9.4. PL BALL COMPLEXES

193

Then there exist PLHs

h'Z

h'1

:

- f' o h'

:

1

An-1

sl(Ki) and

On-1

32(K2')

These induce PLHs 91

and

ga

:

gAn-1 - 8s1()C1I) = 8s1()Ci) 8An-1 - 8s2()Ca) = 8s2()C2)

Now let

B+ := Cl((80+) \ On-1) = (470+) \ int An-1 and

B- := cl((eo-) \ An-1) = (Os-) \ int An-1

B+ and B- are easily seen to be PL-(n - 1) balls. Furthermore, we have 8B+ = 8B- = 80n-1. Hence, using Lemma 9.30, gl and 92 can be extended to PLHs

hi and

hz

: B+ - sl(Ki) : B- -- sl(Ka)

Combining h; and h;' in the obvious way, we get PLHs

hl : 80+ -+ 8s1(Kl) = sl()Ci) U sl()C') ha

: 80- -- 8s2()C2) = s2(K2') U s2(K2)

which agree on An-1. These may again be extended to PLHs h1

h2

A+

- sl(K1) : A- - 32(K2):

and

Now define a complex K whose poset is P and whose space is A+ U 0(which is an n-simplex) in the obvious way: If p E P;, define the space of p to be h; 1s;(p). If p E Pl fl P2, these definitions agree.

The two construction theorems may be put together to yield

Theorem 9.38 (Representation Theorem) Let P be an ndimensional constructible poset. If P is B-constructible there exists a PL-ball complex whose poset is P and whose space is An. If p is S-constructible, there exists a PL-ball complex whose poset is P and whose space is 80n.

194


Proof. This is immediate from the two construction theorems and induction. Note that if P is B-constructible of dimension n and P = [p] with e[p] being S-constructible of dimension n -1, then the inductive assumption gives that there exists a PL-ball complex with poset 8[p] and space 80". In this case, we may simply add p, defining its space to be into" to obtain a complex whose poset is P, and whose space is An. 0

Corollary 9.39 If 0 is the poset of an OM then there exists a PLball complex whose poset is 0 and whose space is

80"+1.

0

From this one easily concludes that OM posets can be represented by sphere systems in the sense of Section 9.3. We present a simple preliminary lemma:

Lemma 9.40 Let 1C = (P, X, s) be a PL-ball complex with s(0) = 0. Then the space of a subcomplex of 1C corresponding to an ndimensional B-(S)-constructible order ideal is a PL n-ball (-sphere).

Proof. Let 1C1 = (P1, X1, sl) be a subcomplex with P1 being Bconstructible of dimension n. By the Representation Theorem there exists a PL-ball complex IC' = (P1, 0", s'). Hence, by the Isomorphism Lemma there exists a PLH : An - X1 showing that X1 is a PL n-ball. The case where P1 is S-constructible is similar. 0

Now let 0 C 2±E be an n-dimensional OM which we may assume to be simple. Let IC = (0, 80"+1, s) be a corresponding PL-ball

complex. For every e E E let HH, H.+, H.-, HI and H: be defined as in Definition 8.39. We claim that s(M) is a hypersphere of s(1C) = 80"+1 with sides s(HH) and s(H; ). To see this, recall from Proposition 8.41 that V.+- and V,-- axe B-constructible with boundary

and s(H

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